I hope these claims are wrong, even though their proofs seem to be the same as the proofs found in textbooks, because I find it very confusing that the minimal assumptions for these results to be true don't seem to actually match the assumptions given.
Note that in what follows I assume all objects are modules over some ring $R$ and the morphisms are module-homomorphisms, and/or assume we are in an abelian category and appeal to the Freyd-Mitchell Embedding Theorem. (I.e. the proofs use elements and not universal constructions.)
The four lemma is commonly stated in terms of two subparts, each with different conclusions (see e.g. here, here, or here).
Note that these two claims seem to be dual to one another, so it would probably be better of me to give an "arrows-only" proof of one, since then the proof of one would imply the other. As far as I am aware, Fredyd-Mitchell does not suffice to conclude that a "non-category-theoretic" proof dualizes.
I had originally hoped the issue was the distinction between the strong and weak four lemmas, i.e. that maybe even though the weak four lemma doesn't require the full exactness assumptions the strong four lemma might require the full exactness assumptions. However the diagram chasing "proofs" below seem to show that this is incorrect, i.e. that the strong four lemma doesn't require the full exactness assumptions usually given either (provided that they're actually correct). This is obviously the least interesting part of my question, so although I think it's important to show that I've given thought to this and have reason to believe these claims are true, the (attempted) proofs distract from and clutters the question itself, hence why it's in a community wiki.
Claim 1$^*$ ("Surjective Part" of Weak Four Lemma): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
which is (E1) exact at $C$, (E2) exact at $B'$, which (C2) is a complex but not necessarily exact at $C'$ (i.e. $h' \circ g' =0$), and neither necessarily exact nor even a complex at $B$, (M1) $\alpha$ is surjective, (M2) $\delta$ is injective, and (M$^*$2) $\gamma$ is surjective, then $\beta$ is surjective.
Claim 2$^*$ ("Injective Part" of Weak Four Lemma): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
which (C1) is a complex but not necessarily exact at $B$ (i.e. $g \circ f = 0$), (E1) exact at $C$, (E2) exact at $B'$, and neither necessarily exact nor even a complex at $C'$, (M1) $\alpha$ is surjective, (M2) $\delta$ is injective, and (M$^*$1) $\beta$ is is injective, then $\gamma$ is injective.
Proof attempt of Claim 1$^*$: Let $b' \in B'$, then because (M$^*$2) $\gamma$ is surjective there exists a $c \in C$ such that $\gamma(c) = g'(b')$. Because (C2) $h' \circ g' = 0$, $h'(\gamma(c)) = h'(g'(b') =0$, and since the rightmost square commutes, it follows that $\delta(h(c))=h'(\gamma(c))=0$. Since (M2) $\delta$ is injective, this means that $h(c)=0$, and therefore that $c \in \operatorname{ker}(h)$, which due to (E1) exactness at $C$ equals $\operatorname{Im}(g)$, thus there exists a $b \in B$ such that $g(b)=c$. Because the middle square commutes, it follows that $g'(\beta(b)) = \gamma(g(b)) = \gamma(c) = g'(b')$, therefore $b' - \beta(b) \in \operatorname{ker}(g')$ which by (E2) exactness at $B'$ is the same as $\operatorname{Im}(f')$, so there exists an $a' \in A'$ such that $f'(a') = b' - \beta(b)$. Moreover, because (M1) $\alpha$ is surjective, there is an $a \in A$ such that $\alpha(a)=a'$, so because the leftmost square commutes, $\beta(f(a))=f'(\alpha(a))=f'(a')= b'-\beta(b)$. Of course since $\beta(f(a)) = b' - \beta(b)$, $b' = \beta(f(a)) + \beta(b) = \beta(f(a)+b)$ for $f(a) + b \in B$. Of course, $b' \in B'$ was arbitrary, so $\beta$ is surjective. $\square$
Proof attempt of Claim 2$^*$: Let $c \in \operatorname{ker}(\gamma)$, then $h'(\gamma(c)) = h'(0)=0$, and thus since the rightmost square commutes, $\delta(h(c))=h'(\gamma(c))=0$, and since (M2) $\delta$ is injective, $\delta(h(c))=0$ means that $h(c)=0$, so $c \in \operatorname{ker}(h)$ which (E1) equals $\operatorname{Im}(g)$ due to exactness at $C$. Thus there exists a $b \in B$ such that $g(b)=c$, and thus since the middle square commutes, $g'(\beta(b)) = \gamma(g(b))=\gamma(c)=0$, which means that $\beta(b) \in \operatorname{ker}(g')$, which by (E2) exactness at $B'$ equals $\operatorname{Im}(f')$, therefore there exists an $a' \in A'$ such that $f'(a') = \beta(b)$. Since (M1) $\alpha$ is surjective, there is an $a \in A$ such that $\alpha(a) = a'$, and therefore since the leftmost square commutes, $\beta(f(a)) = f'(\alpha(a))f'(a') = \beta(b)$. Since (M$^*$1) $\beta$ is injective, this means that $f(a)=b$. Therefore, since (C1) $g \circ f =0$, it follows that $c = g(b) = g(f(a)) = 0$. Since $c \in \operatorname{ker}(\gamma)$ was arbitrary, it follows that $\operatorname{ker}(\gamma)=\{0\}$, i.e. that $\gamma$ is injective. $\square$
Corollary (Full Weak Four Lemma Under Weakened Assumptions): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
which (E1) is exact at $C$, (E2) exact at $B'$, (C1) a complex at $B$ (i.e. $g \circ f =0$), (C2) a complex at $C'$ (i.e. $h' \circ g' = 0$), (M1) $\alpha$ is surjective, and (M2) $\delta$ is injective, then both parts of the weak four lemma are simultaneously true.
Claim 1 ("Image Part" of Strong Four Lemma): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
where (M1) $\alpha$ is surjective, (M2) $\delta$ is injective, which is (E1) exact at $C$ and (E2) exact at $B'$, which is (C2) a complex but not necessarily exact at $C'$ (i.e. $h' \circ g' = 0$), and neither necessarily exact nor even a complex at $B$, then $\operatorname{Im}(\beta) = (g')^{-1}(\operatorname{Im}(\gamma)) $.
Proof attempt of Claim 1: First we note the much more general claim that
Claim 1(a): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
B @>g>> C \\
@V\beta VV @V\gamma VV \\
B' @>g'>> C' \\
\end{CD}
$$
one always has that $\operatorname{Im}(\beta) \subseteq (g')^{-1}(\operatorname{Im}(\gamma))$.
Let $b' \in \operatorname{Im}(\beta)$, then $b' = \beta(b)$ for some $b \in B$, and since the diagram commutes, $g'(b')=g'(\beta(b))=\gamma(g(b))= \gamma(c)$ for a $c=g(b)\in C$. Therefore $b' \in (g')^{-1}(\operatorname{Im}(\gamma))$. $\triangle$
In light of Claim 1(a), we see that the only important substance of Claim 1 consists in proving the inclusion $(g')^{-1}(\operatorname{Im}(\gamma)) \subseteq \operatorname{Im}(\beta)$.
So let $b' \in (g')^{-1}(\operatorname{Im}(\gamma))$, then there exists $c \in C$ such that $g'(b') = \gamma(c)$. Then because (C2) $h' \circ g' = 0$, $h'(\gamma(c))= h'(g'(b'))=0$, and since the rightmost square commutes, $\delta(h(c))=h'(\gamma(c))=0$, and since (M2) $\delta$ is injective, this means that $h(c)=0$, and thus that $c \in \operatorname{ker}(h)$. Since (E1) the diagram is exact at $C$, $\operatorname{ker}(h)=\operatorname{Im}(g)$, so there exists a $b \in B$ such that $g(b)=c$. Note that since the middle square commutes, $g'(\beta(b))=\gamma(g(c))=g'(b')$ (by definition of $c$), therefore $b ' - \beta(b) \in \operatorname{ker}(g')$. By (E2) exactness at $B'$, $\operatorname{ker}(g')=\operatorname{Im}(f')$, so there exists an $a' \in A'$ such that $f'(a') = b' - \beta(b)$, and since (M1) $\alpha$ is surjective, there is an $a \in A$ such that $\alpha(a)=a'$. Then because the leftmost square commutes, $b' - \beta(b) = f'(a')=f'(\alpha(a))=\beta(f(a))$, which means that $b' = \beta(b)+\beta(f(a)) = \beta(b + f(a))$, which means that $b' \in \operatorname{Im}(\beta)$. Of course, $b' \in (g')^{-1}(\operatorname{Im}(\gamma))$ was arbitrary, so $(g')^{-1}(\operatorname{Im}(\gamma)) \subseteq \operatorname{Im}(\beta)$. $\square$
Claim 2 ("Kernel Part" of Strong Four Lemma) Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
where (M1) $\alpha$ is surjective, (M2) $\delta$ is injective, which (E1) is exact at $C$, (E2) is exact at $B'$, which is (C1) a complex but not necessarily exact at $B$ (i.e. $g \circ f = 0$), and neither necessarily exact nor even a complex at $C'$, then $g(\operatorname{ker}(\beta))=\operatorname{ker}(\gamma)$.
Proof attempt of Claim 2: First we note the much more general claim that
Claim 2(a): Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
B @>g>> C \\
@V\beta VV @V\gamma VV \\
B' @>g'>> C' \\
\end{CD}
$$
one always has that $g(\operatorname{ker}(\beta)) \subseteq \operatorname{ker}(\gamma)$.
Let $c \in g(\operatorname{ker}(\beta))$, then $c = g(b)$ for a $b \in \operatorname{ker}(\beta)$, so since the diagram commutes, $\gamma(c)=\gamma(g(b))=g'(\beta(b))=g'(0)=0$, so $c \in \operatorname{ker}(\gamma)$. $\triangle$
In light of Claim 2(a), we see that the only important substance of Claim 2 consists in proving the inclusion $\operatorname{ker}(\gamma) \subseteq g(\operatorname{ker}(\beta))$.
So let $c \in \operatorname{ker}(\gamma)$. Then because the rightmost square commutes, $\delta(h(c)) = h'(\gamma(c))= h'(0)=0$, and since (M2) $\delta$ is injective, this means that $h(c)=0$, thus that $c \in \operatorname{ker}(h)$. By (E1) exactness at $C$, $\operatorname{ker}(h)=\operatorname{Im}(g)$, so there exists a $b \in B$ such that $g(b)=c$. Because the middle square commutes, $g'(\beta(b))=\gamma(g(b))=\gamma(c)=0$, thus $\beta(b) \in \operatorname{ker}(g')$, and since by (E2) exactness at $B'$, i.e. $\operatorname{ker}(g')=\operatorname{Im}(f')$, there exists an $a' \in A'$ such that $f'(a') = \beta(b)$. Because (M1) $\alpha$ is surjective, there is an $a\in A$ such that $\alpha(a)=a'$, so because the leftmost square commutes, $\beta(f(a))=f'(\alpha(a))=f'(a')=\beta(b)$. Therefore $b - f(a) \in \operatorname{ker}(\beta)$. Note that by (C1) $g(b - f(a))= g(b) - g(f(a)) = c - 0 = c$, thus $c = g(\tilde{b})$ for a $\tilde{b} \in \operatorname{ker}(\beta)$, i.e. $c \in g(\operatorname{ker}(\beta))$. Of course, since $c \in \operatorname{ker}(\gamma)$ was arbitrary, $\operatorname{ker}(\gamma) \subseteq g(\operatorname{ker}(\beta))$. $\square$
Corollary (Full Strong Four Lemma Under Weakened Assumptions) Given the commutative diagram
$$
\require{AMScd}
\begin{CD}
A @>f>> B @>g>> C @>h>> D \\
@V\alpha VV @V\beta VV @V \gamma VV @V \delta VV \\
A' @>f'>> B' @>g'>> C' @>h'>> D' \\
\end{CD}
$$
such that the following assumptions hold:
- (M1) $\alpha$ is surjective
- (M2) $\delta$ is injective
- (E1) the diagram is exact at $C$
- (E2) the diagram is exact at $B'$
- (C1) the diagram is a complex at $B$ (i.e. $g\circ f = 0$)
- (C2) the diagram is a complex at $C'$ (i.e. $h' \circ g' = 0$)
then:
- ("Image Part") $\operatorname{Im}(\beta) = (g')^{-1}(\operatorname{Im}(\gamma)) $, $\quad$ and
- ("Kernel Part") $g(\operatorname{ker}(\beta)) = \operatorname{ker}(\gamma)$
