Assume an arbitrary periodic function $f(t)$ with period $T$. Consider the Fourier series representation of $f(t)$, in which $\omega_0=\frac{2\pi}{T}$:
$$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}$$
Take the Fourier transform of the sides:
\begin{align}
\mathcal{F}\{f(t)\}=&\mathcal{F}\{\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}\}\\
=&\sum_{n=-\infty}^{+\infty}c_n\mathcal{F}\{ e^{i n \omega_0 t}\}\\
F(\omega)=&2\pi\sum_{n=-\infty}^{+\infty}c_n\delta(\omega-n\omega_0)
\end{align}
This means that the Fourier transform of a periodic signal is an impulse train where the impulse amplitudes are $2\pi$ times the Fourier coefficients of that signal.
With $f(t)=\delta(t)$, the Fourier series coefficients are $c_n=\frac{1}{T}$ for all $n$.
Hence,
$$\mathcal{F}\{\sum_{n=-\infty}^{+\infty}\delta(t-nT)\}=\frac{2\pi}{T} \sum_{n=-\infty}^{+\infty}\delta(\omega-n\omega_0)$$
or in comb notation:
$$\boxed{\mathcal{F}\{\text{comb}_T(t)\}=\omega_0\ \text{comb}_{\omega_0}(\omega)}$$
where
$$\text{comb}_A(x)\triangleq\sum_{n=-\infty}^{+\infty}\delta(x-nA)$$
Let us define the Fourier Transform of a function $f\in\mathcal{S}(\mathbb R)$ by
$$\hat{f}(x)=c\int_\mathbb Rf(y)e^{-iyx}\;dy,$$
where $c=\frac{1}{\sqrt{2\pi}}$. You are okay with the following result.
(1) Inversion formula in $\mathcal{S}(\mathbb R)$. If $f\in\mathcal{S}(\mathbb R)$, then
$$f(x)=c\int_\mathbb R\hat{f}(y)e^{iyx}\;dy,\quad\forall\ x\in\mathbb {R}.$$
You want to pass to $L^1$. In other words, you want to prove the following result.
(2) Inversion formula in $L^1$. If $f\in L^1$ and $\hat{f}\in L^1$, then
$$f(x)=c\int_\mathbb R\hat{f}(y)e^{iyx}\;dy,\quad \text{a.e.}\ x\in\mathbb {R}.$$
You know that you should use the inclusions $C^{\infty}_0 \subset S(\mathbb{R}) \subset L^{1}$ but don't know how. Here is some details, from Hörmmander's book, p. 164 (which proves the result for functions in $\mathcal{S}'(\mathbb R)$, the space of tempered distributions).
Proof of (2): From (1), given $\varphi\in\mathcal S(\mathbb R)$,
$$\varphi(-x)=c\int_\mathbb R\hat{\varphi}(y)e^{iy(-x)}\;dy=\hat{\hat{\varphi}}(x),\quad\forall\ x\in\mathbb {R}.$$
So,
$$\int_\mathbb R \hat{\hat{f}}(x)\varphi(x)\;dx\overset{(*)}{=}\int_\mathbb R f(x)\hat{\hat{\varphi}}(x)\;dx=\int_\mathbb R f(x)\varphi(-x)\;dx=\int_\mathbb R f(-x)\varphi(x)\;dx,\quad\forall\ \varphi\in C_0^\infty$$
and thus $\hat{\hat{f}}(x)=f(-x)$ a.e. $x\in\mathbb R$ (by the du Bois Reymond Lemma), which implies
$$f(x)=\hat{\hat{f}}(-x)=c\int_\mathbb R \hat{f}(y)e^{-iy(-x)}\;dy=c\int_\mathbb R \hat{f}(y)e^{iyx}\;dy,\quad\text{a.e. } x\in \mathbb R.$$
Note that the equality $(*)$ is valid because $f,\hat{f},\varphi,\hat{\varphi}\in L^1$. $\square$
Remark 1: The inversion formulas (1) and (2) allows us to define the Inverse Fourier Transform in $\mathcal{S}(\mathbb R)$ and $L^1$. In $L^2$, the Inverse Fourier Transform (as well as the the Fourier Transform) is defined by extension. So, it is not clear for me what you call "inversion formula in $L^2$".
Remark 2: In $\mathbb R^n$, the same argument works.
Best Answer
Well, you could show that for $f\in C_c(\mathbb{R})$, $\lvert\hat{f}(\xi)\rvert=O(\lvert\xi\rvert^{-1})$ as $\lvert\xi\rvert\to\infty$, and similarly for $C^1_c(\mathbb{R})$ you then get $O(\lvert\xi\rvert^{-2})$. This gives $\hat{f}\in L^1$.
Of course you would recognize it as part of the proof of Fourier transform of Schwartz is Schwartz, adapted to our case.