Let us, to distinguish between the two Fourier transforms, denote the one by $\mathscr{F}_1$,
$$\mathscr{F}_1(f)(y) = \int_\mathbb{R} f(x)e^{-ixy}\,dx\tag{1}$$
for $f\in L^1(\mathbb{R})$, and the other by $\mathscr{F}_2$,
$$\mathscr{F}_2(g) = \lim_{n\to\infty} \mathscr{F}_1(g_n)\tag{2}$$
for $g\in L^2(\mathbb{R})$ where $g_n$ is a sequence of functions in $S$ that converges to $g$ in $L^2$, and the limit in $(2)$ is the $L^2$-limit.
For $f \in L^1 \cap L^2$, find a sequence $(g_n)$ in $S$ such that not only $\lVert g_n - f\rVert_2 \to 0$ but also $\lVert g_n - f\rVert_1 \to 0$.(1)
Then you can conclude
$$\left\lVert \mathscr{F}_1(g_n) - \mathscr{F}_1(f)\right\rVert_\infty \leqslant \lVert g_n - f\rVert_1 \to 0,$$
i.e. the Fourier transforms of the $g_n$ converge uniformly to the $L^1$ Fourier transform of $f$, and by definition you have
$$\mathscr{F}_1(g_n) \xrightarrow{L^2} \mathscr{F}_2(f),$$
and there is a subsequence of the $\mathscr{F}_1(g_n)$ that converges pointwise almost everywhere to $\mathscr{F}_2(f)$ (well, we know it converges uniformly, so the entire sequence converges pointwise), so you have $\mathscr{F}_1(f) = \mathscr{F}_2(f)$ almost everywhere (and you can choose $\mathscr{F}_1(f)$ as a representative, so then you have equality everywhere). In that sense, the two definitions coincide, for $f \in L^1\cap L^2$, the $L^1$-Fourier transform of $f$ is a representative of the $L^2$-Fourier transform of $f$.
If you choose a sequence $(g_n)$ in $S$ such that $\lVert g_n - f\rVert_2 \to 0$, but not $\lVert g_n -f \rVert_1 \to 0$, then by the above you have convergence $\hat{g}_n \xrightarrow{L^2} \hat{f}$, and that implies that for a subsequence $g_{n_k}$, you have pointwise convergence $\hat{g}_{n_k}(y) \to \hat{f}(y)$ almost everywhere, but there is no guarantee (known to me) that you have pointwise a.e. convergence for the full sequence, nor that you have pointwise convergence everywhere for any subsequence. But it would be difficult at least, I think, to come up with a concrete example of a function $f\in L^1\cap L^2$ and a sequence $g_n$ of Schwartz functions converging to $f$ in $L^2$ such that you don't have pointwise convergence almost everywhere or even everywhere for the full sequence.
(1) That is always possible, let $f_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x)$. Then $f_m \to f$ both in $L^1$ and $L^2$. $f_m$ is a bounded function with compact support, so convolving it with a compactly supported approximation of the identity produces compactly supported smooth functions converging to $f_m$ in both $L^1$ and $L^2$.
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
My outline of solution:
Replacing $\hat{g}(y)$ by the integral, we will obtain $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{ixy}dy= \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(z)e^{iy(x-z)}dz\right)dy$$
Next we multiply a term $e^{-\frac{\epsilon^2y^2}{4}}$ into the integration, and consider the limit as $\epsilon\to0$,
$$A(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(z)e^{-\frac{\epsilon^2y^2}{4}}e^{iy(x-z)}dz\right)dy$$ After simplifying, we get $$A(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}dy$$ Since $|\hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}|\leq|\hat{g}(y)|$, $$\lim_{\epsilon\to0}A(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\lim_{\epsilon\to0} \hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}dy=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{ixy}dy$$...