Let's say I have the function $g(x) = \exp(-x)1_{[0,\infty)}$ and I would like to find another function $f$, whose fourier transform is $g$.
Since the fourier inversion theorem only works for $f \in L^1$, I would first have to show, that the original function is indeed integrable. I know that on $L^1(\mathbb R)$ the fourier transform is a continuous linear map to $C_0(\mathbb R)$, but since $g$ is obviously not continuous at $0$, I would argue that $f$ is not in $L^1$.
However, if I nonetheless try to use the inversion theorem, I get $$f= \frac{1}{\sqrt{2\pi}}\frac{1}{ix+1} $$
I let Wolframalpha compute the fourier transform of $f$ and I seem to be getting back my original function. Since I am not really familiar with complex analysis or distribution theory, this result confuses me a little bit. It seems like $f$ is not in $L^1$, but is in $L^2$. Was the usage of the inversion theorem justified here in any way? (I am familiar with the Fourier-Plancherel operator, so my intuition here was, that I just got lucky, but applying Fourier-Plancherel to our $f$, gives us back $g$)
It is not just being lucky.
Your original function $g$ belongs to $L^2$ (it is actually in $L^1 \cap L^2)$. This means that its FT is an $L^2$ function and the inversion formula holds.
Extended
Given a function $f \in L^1 \cap L^2$, by Plancharel Theorem we have $$\| f \|_2 = \|\widehat{f}\|_2$$
It follows that the FT is an isometry from $L^1 \cap L^2$ to $L^2$. Since $L^1 \cap L^2$ is dense in $L^2$ it follows that the FT has an unique extension to a linear operator $\widehat{ \cdot } : L^2 \to L^2$.
Now, as the Fourier Inversion Formual holds on a dense subspace of $L^2$, it follows that for all $f \in L^2$ we have $$\widehat{\hat{f}}(x)=f(-x)$$
But, be carefull. For $f \in L^1 \cap L^2$, the Fourier trasnform $\widehat{f}$ has classical meaning. If $f \in L^2$ is not integrable, the meaning of $\widehat{f}$ is $$\widehat{f}\stackrel{\| \, \|_2}{=}\lim_n \widehat{f_n} \\ \mbox{where} f_n \in L^1 \cap L^2 \, \mbox{ and } f\stackrel{\| \, \|_2}{=}\lim_n f_n$$
The discussion at the top explains why this definition does not depend on the choice of $f_n$.