Rephrasing the question, let $A, B \in \operatorname{Mat}_3(\mathbb{C})$ be two invertible matrices such that $AB = BA$, and $A - B$ and $AB - A$ are skew-symmetric. Is it true that 1 is always an eigenvalue of $AB$?
Since $A - B$ is skew-symmetric, we may write $A = Q + S$ and $B = Q + T$ for some symmetric matrix $Q$ and skew-symmetric matrices $S, T$. Applying the condition $AB = BA$ gives
$$\begin{aligned}
0
&= AB - BA\\
&= (Q^2 + QT + SQ + ST) - (Q^2 + QS + TQ + TS) \\
&= \underbrace{(SQ - QS)}_{\text{symmetric}} + \underbrace{(QT - TQ)}_{\text{symmetric}} + \underbrace{(ST - TS)}_{\text{skew-symmetric}}.
\end{aligned}$$
Equating parts gives $ST = TS$ and $(S-T)Q = Q(S-T)$, or in other words $S$ commutes with $T$ and $Q$ commutes with $(S-T)$.
Now looking at the next condition, we have
$$\begin{aligned}
AB - A
&= (Q^2 + QT + SQ + ST) - (Q + S) \\
&= \underbrace{(Q^2 - Q + ST)}_{\text{symmetric}} + \underbrace{(QT + SQ - S)}_{\text{skew-symmetric}},
\end{aligned}$$
where we can see that $QT + SQ$ is skew-symmetric by rearranging $(S-T)Q=Q(S-T)$. Since $AB - A$ should be purely skew-symmetric, we find $ST = Q - Q^2$. In particular, this also shows that $Q$ commutes with $ST$.
Now suppose that one of $S$ or $T$ is zero, say $S = 0$. The equation $ST = Q - Q^2$ implies that $Q = Q^2$, and since $A = Q$ is invertible we can cancel to get $Q = I$. Then $AB = I + T$ has 1 as an eigenvalue, since we can take any vector $v$ in the kernel of $T$ (which necessarily exists, since $T$ is skew-symmetric on an odd-dimensional space) and get $ABv = (I +T)v = v$.
Suppose from now on that $S \neq 0$ and $T \neq 0$, each with eigenvalues $\{\pm s, 0\}$ and $\{\pm t, 0\}$ for $s, t \neq 0$. Since $S$ and $T$ commute, and have distinct eignevalues, they simultaneously diagonalise. We have essentially two possibilities for what these diagonalisations look like, depending on whether $\ker ST$ is one-dimensional or two-dimensional. (I'm using the shorthand $[S]$ to mean $P^{-1} S P$ for the simultaneous diagonalising matrix $P$).
Case 1:
$$\begin{aligned}\,
[S] = \begin{pmatrix}s & & \\ & -s & \\ & & 0 \end{pmatrix}, \quad
[T] = \begin{pmatrix}t & & \\ & -t & \\ & & 0 \end{pmatrix}, \quad
[ST] = \begin{pmatrix}st & & \\ & st & \\ & & 0 \end{pmatrix}, \quad
[S-T] = \begin{pmatrix}s-t & & \\ & t-s & \\ & & 0 \end{pmatrix}
\end{aligned}$$
Case 2:
$$\begin{aligned}\,
[S] = \begin{pmatrix}s & & \\ & -s & \\ & & 0 \end{pmatrix}, \quad
[T] = \begin{pmatrix}t & & \\ & 0 & \\ & & -t \end{pmatrix}, \quad
[ST] = \begin{pmatrix}st & & \\ & 0 & \\ & & 0 \end{pmatrix}, \quad
[S-T] = \begin{pmatrix}s-t & & \\ & -s & \\ & & t \end{pmatrix}
\end{aligned}$$
In case 1, let $v \in \ker ST$, necessarily we also have $Sv = Tv = 0$. Since $Q$ commutes with $ST$ and $\ker ST$ is one-dimensional, $v$ is also an eigenvector of $Q$, say $Qv = qv$. Applying $ST = Q - Q^2$ to $v$ gives $q^2 - q = 0$, and since $Sv = 0$ we have $Av = Qv \neq 0$ since $A$ is invertible. Hence $q = 1$. Then $ABv = Qv = v$, so $AB$ has 1 as an eigenvalue.
In case 2 we can't use the same approach, since a vector in $\ker ST$ is not necessarily an eigenvector of $Q$, due to the kernel being two-dimensional. However we can instead use $S - T$ to show we have an appropriate eigenvector. Since $S - T$ is skew-symmetric it must have zero as an eigenvalue, and since by assumption $s, t \neq 0$ we must have $s = t$, i.e. the top-left entry of $S - T$ is zero. This means that $S - T \neq 0$, and since it commutes with $Q$, $Q$ also simultaneously diagonalises in the same basis as $S$ and $T$. We can then choose $v \in \ker S$ again (or $\ker T$) and repeat similar logic to show that $AB$ has 1 as an eigenvalue.
