Not related to the body of the post, but rather the title.
One can provide an explicit example to show the Fourier transform is not surjective: https://mathoverflow.net/a/319225/112504 proves that $$g(t) := 1_{[0\leq t\leq e]}\cdot \frac te + 1_{[t>e]}\cdot \frac{1}{\ln t},$$ extended to be an odd continuous function on $\mathbb R$ can not be the Fourier transform of any $L^1(\mathbb R)$ function. That answer cites: "this example is taken from the book Classical Fourier Transforms, Springer–Verlag,
1989 by K. Chandrasekharan." For the convenience of the reader, I will copy and paste the proof from the answer's source (M. Thamban Nair's Fourier Analysis notes):
Proposition 3.2.1 If $f \in L^1(\mathbb{R})$ is an odd function, then $\hat{f}$ is an odd function and there exists $M>0$ such that $$
\left|\int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi\right| \leq M $$ for all
$r, R$ with $0<r<R$. Proof. Let $f \in L^1(\mathbb{R})$ is an odd function. It can be easily seen that ${ }^4$, $$ \hat{f}(\xi)=2 i \int_0^{\infty} f(x) \sin (\xi x) d x $$
Thus, $\hat{f}$ is odd. Let $R \geq r>0$. Then, we have $$
\begin{aligned} \int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi & =2 i
\int_r^R \frac{1}{\xi}\left(\int_0^{\infty} f(x) \sin (\xi x) d
x\right) d \xi \\ & =2 i \int_0^{\infty} f(x)\left(\int_r^R \frac{\sin
(\xi x)}{\xi} d \xi\right) d x \\ & =2 i \int_0^{\infty}
f(x)\left(\int_{r x}^{R x} \frac{\sin (s)}{s} d s\right) d x
\end{aligned} $$
We know that there exists $M_0>0$ such that $\left|\int_a^b \frac{\sin
x}{x} d x\right| \leq M_0$ for all $(a, b) \subseteq \mathbb{R}$.
Thus, $$ \left|\int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi\right| \leq 2
\int_0^{\infty}|f(x)|\left|\int_{r x}^{R x} \frac{\sin (s)}{s} d
s\right| d x \leq 2 M_0\|f\|_1 . $$
$\blacksquare$
Theorem 3.2.2 The bounded linear operator $f \mapsto \hat{f}$ is a bounded linear operator from $L^1(\mathbb{R})$ to
$C_0\left(\mathbb{R}\right)$ is not onto. Proof. By Proposition 3.2.1, it is enough to construct an odd function $g \in C_0(\mathbb{R})$ such that $$ \left|\int_r^R \frac{g(t)}{t} d t\right| \rightarrow \infty \quad \text { as } \quad R \rightarrow \infty . $$
A candidate for such a function is the odd extension of $g$ defined by $$ g(t):= \begin{cases}t / e, & 0 \leq t \leq e \\ 1 / \ln (t), & t>e\end{cases} $$
Note that $$ \int_e^R \frac{g(t)}{t} d t=\ln (\ln (R)) \rightarrow \infty \quad \text { as } \quad R \rightarrow \infty $$
$\blacksquare$
$\color{red}{\text{That example was based on a $C_0(\mathbb R)$ function decaying slowly to $0$.}}$
Here seems to be one example of a compactly supported continuous function who is not the Fourier transform of any $L^1(\mathbb R)$ function: https://mathoverflow.net/a/95678/112504 (it suffices to find an example of a $g\in C_c(\mathbb R)$ function whose Fourier transform $\hat g$ isn't in $L^1(\mathbb R)$, since if we assume for sake of contradiction $g = \hat f$ for some $f\in L^1(\mathbb R)$, then Fourier inversion would tell us that $\hat g(s)=f(-s) \in L^1(\mathbb R)$ (perhaps up to constant factors); contradiction)
Choose
$$g(x)=
\begin{cases}
\dfrac{\frac12 -x}{\log(x)},&0<x\leq1/2\\
0,&\text{otherwise}
\end{cases}$$
Here's a graph of this function for reference
Here's the proof: assume for sake of contradiction that $\hat g$ is in $L^1$, so all the integrals in the following line are well-defined:
$$\begin{aligned}
\infty > \|\hat g\|_{L^1(\mathbb R)} &= \int_{-\infty}^\infty |\hat g(s)| d s \geq \int_0^\infty |\hat g(s)| d s \geq \int_0^\infty \lvert\operatorname{Im} \hat g(s)\rvert ds \\
&\geq \left|\int_0^\infty \operatorname{Im} \hat g(s)ds\right| \geq \int_0^\infty \operatorname{Im} \hat g(s)ds
\end{aligned}$$
In fact, because for $s\in [0,\infty)$, $e^{-\varepsilon s} \operatorname{Im}\hat g(s) $ is dominated $|e^{-\varepsilon s} \operatorname{Im}\hat g(s)|\leq |\hat g(s)|$ we we assumed was in $L^1(\mathbb R_s)$, the DCT (dominated convergence theorem) tells us that
$$\lim_{\varepsilon \searrow 0} \int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds = \int_0^\infty \operatorname{Im} \hat g(s)ds.$$
Note that
$$\operatorname{Im} \hat g(s) = \int_{-\infty}^\infty g(x) \operatorname{Im}(e^{-ix s}) dx = \int_{0}^{1/2} g(x) \sin(-xs) dx.$$
But
$$\begin{aligned}
\int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds &= \int_0^\infty e^{-\varepsilon s} \int_0^{1/2} (-g(x)) \sin(xs) dx ds \\
&= \int_0^{1/2} (-g(x)) \int_0^\infty e^{-\varepsilon s} \sin (xs) ds dx.
\end{aligned}$$
where we used Fubini-Tonelli, which we can use once we check the Tonelli condition:
$$ \int_0^{1/2} \int_0^\infty |e^{-\varepsilon s} (-g(x)) \sin(xs)| ds dx \leq \int_0^{1/2} \underbrace{|g(x)|}_{\leq 0.2} \underbrace{\int_0^\infty e^{-\varepsilon s} ds}_{=\frac 1{\varepsilon} e^{-\varepsilon}} dx < \infty$$
Finally, it is an easy exercise (using Euler's identity and taking Re/Im parts, or twice integration by parts) to see that
$$\int_0^\infty e^{-\varepsilon s} \sin (xs) ds = \lim_{R\nearrow \infty} \left.\left(\frac{e^{-\varepsilon s}}{\varepsilon^2+x^2} (-\varepsilon \sin(xs) - x \cos (xs) ) \right)\right|_{s=0}^{s=R} = \frac{x}{\varepsilon^2+x^2}$$
and so (using that $-g(x)=|g(x)|\geq -\frac 14\cdot \frac{1}{\log x} \cdot 1_{[0,\frac 14]}$)
$$\int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds = \int_0^{1/2} (-g(x)) \cdot \frac{x}{\varepsilon^2+x^2} dx \geq \int_0^{1/4} \frac 14\cdot \frac{1}{-\log x} \cdot \frac{x}{\varepsilon^2+x^2} dx.$$
As $\varepsilon\searrow 0$, the RHS goes to $\int_0^{1/4} \frac 14 \frac{1}{-x\log x} dx$ by MCT (monotone convergence theorem), which equals $\infty$; tracing back all our steps, we get that $\infty > \|\hat g \|_{L^1(\mathbb R)}\geq \infty$; contradiction.
$\color{red}{\text{The key feature that made this chosen function work is its steepness at $0$.}}$ It is only by having such a steep transition from constant $0$ on $(-\infty,0]$ to something nonzero, that $(-g(x)) \cdot \frac 1x$ preserves the non-$L^1$-ness of $\frac 1x$ at $0$.
One corollary of the above example is that one can have a sequence of very nice functions in $C_c(\mathbb R)$ (all of which come from Fourier transforms of very nice functions) converging very nicely to a function in $C_c(\mathbb R)$ which is not the Fourier transform of any $L^1(\mathbb R)$ function.
More precisely, one can take a sequence $g_n\in C_c^\infty ([0,\frac 12])$ converging in the sup-norm (and hence in all the $L^p$-norms) to $g \in C_c([0,\frac 12])$; all the $g_n$ (by Fourier inversion on $C_c^\infty \subseteq \mathcal S$) come from Schwartz functions $f_n\in \mathcal S(\mathbb R) \subseteq L^1(\mathbb R)$, but there is no $f\in L^1(\mathbb R)$ s.t. $\hat f = g$.
So the $f_n$ are infinitely smooth everywhere, faster than any polynomial decay (in $C_0^\infty(\mathbb R)$! in every $L^p(\mathbb R)$ space!), and the $g_n$ are infinitely smooth and compactly supported! And the $g_n$ converge in "every possible way" to $g$! And yet, we can't even find an integrable $f$ whose Fourier transform is $g$.
P.S. more examples can be found elsewhere on MSE. I've tried to collect those that I could find here: A Fourier transform of a continuous $L^1$ function (gives an example quite similar to my 2nd example above, the one with compact support)
Some questions I have remaining:
