That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.
Just to establish where we're putting the $\pi$'s, we define $$\hat f(\xi)=\int f(t)e^{-it\xi}\,dt.$$
$L^1$ Inversion Theorem. If $f\in L^1(\Bbb R)$ and $\hat f\in L^1(\Bbb R)$ then $f(t)=\frac1{2\pi}\int\hat f(\xi)e^{i\xi t}\,d\xi$ almost everywhere.
We use that periodization argument to establish the theorem under stronger hypotheses:
Partial Inversion Theorem. If $f,f',f''\in L^1(\Bbb R)$ then $\hat f\in L^1$ and $f(t)=\frac1{2\pi}\int\hat f(\xi)e^{it\xi}\,d\xi$.
To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''\in L^1$.
Note first that $(1+\xi^2)\hat f(\xi)$ is the Fourier transform of $f-f''$ (see Details below), so it's bounded: $$|\hat f(\xi)|\le\frac c{1+\xi^2}.\tag{*}$$
For $L>0$ define $$f_L(t)=\sum_{k\in\Bbb Z}f(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_{L,n}=\frac1L\int_0^Lf_L(t)e^{-2\pi i n t/L}\,dt.$$
Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_{L,n}=\frac1L\hat f\left(\frac{2\pi n}L\right).$$So ($*$) above shows that $\sum_n|c_{L,n}|<\infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=\frac1L\sum_n\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$\frac1L\sum_n\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}=\frac1{2\pi}\int g_L(\xi)\,d\xi,$$where $$g_L(\xi)=\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}\quad(\xi\in[2\pi n/L,2\pi(n+1)/L)).$$Since $\hat f$ is continuous, DCT (using ($*$) for the D) shows that $$\lim_{L\to\infty}\int g_L=\int\hat f(\xi)e^{i\xi t}\,d\xi.$$
So we're done if we can show that $f_L\to f$ almost everywhere as $L\to\infty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_L\to f$ in $L^1_{loc}$ for every $f\in L^1$, hence some subsequence tends to $f$ almost everywhere.
Deriving IT from PIT is very simple. Say $(\phi_n)$ is an approximate identity; in particular $\phi_n\in C^\infty_c$, the support of $\phi_n$ shrinks to the origin, $||\phi_n||_1=1$ and $\hat\phi_n\to1$ pointwise. Let $f_n=f*\phi_n$. Then $f_n'=f*\phi_n'$, so $f'\in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_n\to f$ almost everywhere and DCT shows that $||\hat f_n-\hat f||_1\to0$.
Details, in answer to a comment. Note that here when I say $f,f'\in L^1$ I mean that $f$ is absolutely continuous and $f'\in L^1$.
Proposition. If $f,f'\in L^1(\Bbb R)$ then $\widehat{f'}(\xi)=-i\xi\hat f(\xi)$.
(Unless it's $i\xi\hat f(\xi)$; I never remember - here it doesn't matter since $(-1)^2=1$.)
Of course the proposition is just an integration by parts. Then we have to justify integration by parts in this context and worry about the boundary terms. Seems more instructive to show that
Given $f\in L^1$, the following are equivalent: (i) $f'\in L^1$, (ii) $f$ is "differentiable in $L^1$".
Regarding what (ii) means, see Lemma 2 below. I like to go this way because first, it's cute: "$f'\in L^1$ if and only if $f$ is differentiable in $L^1$", and second it seems to me to say something about what absolute continuity "really means". Anyway:
Exercise. If $f\in L^1$ then $\lim_{t\to0}\int|f(x)-f(x+t)|\,dx=0$.
(Hint: Wlog $f\in C_c(\Bbb R)$.)
Lemma 1. If $f\in L^1$ then $\lim_{h\to0}\int\left|f(x)-\frac1h\int_x^{x+h}f(t)\,dt\right|\,dx=0$.
Proof: $$\begin{align}\int\left|f(x)-\frac1h\int_x^{x+h}f(t)\,dt\right|\,dx
&=\int\left|\frac1h\int_0^h(f(x)-f(x+t))\,dt\right|\,dx
\\&\le\frac1h\int_0^h\int|f(x)-f(x+t)|\,dxdt.\end{align}$$
Apply the previous exercise and note that $\frac1h\int_0^h\epsilon=\epsilon$.
Lemma 2. If $f,f'\in L^1$ then $\lim_{h\to0}\int\left|f'(x)-\frac{f(x+h)-f(x)}{h}\right|\,dx=0$.
That is, if $f,f'\in L^1$ then $f$ is "differentiable in $L^1$". (We won't use the other implication...)
Proof: Write $\frac{f(x+h)-f(x)}{h}=\frac1h\int_x^{x+h}f'(t)\,dt$ and apply Lemma 1.
Another interesting/instructive version of "differentiable in $L^1$" that we won't use below:
Exercise. Suppose $f\in L^1$, and define $F:\Bbb R\to L^1(\Bbb R)$ by $F(t)(x)=f(x+t)$. Then (i) $f'\in L^1$ if and only if (ii) $F$ is differentiable.
Proof of the proposition: Work out the Fourier transform of the function $x\mapsto\frac{f(x+h)-f(x)}{h}$. Let $h\to0$ (apply Lemma 2).
Best Answer
No you cannot, Parseval's Identity works in Hilbert spaces, e.g. $L^2$ but not in a general Banach space e.g. $L^1$.