Truth of the Poisson summation formula
In what follows, I'll use the convention $$ \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i x \xi}dx,$$ so that $$ f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi)e^{2\pi i x \xi}d\xi.$$
I like the following interpretation of Poisson summation, which also gives a generalization: Consider the Dirac comb distribution $C(x) = \sum_{n\in \mathbb{Z}} \delta(x-n)$. This is a tempered distribution, so it has a Fourier transform. In fact, it is its own Fourier transform. To justify this, I'm going to give a very nonrigorous argument. But if intuition is the main goal, then I think it will help. First, note that $C(x)$ is periodic with period 1. Thus, its "Fourier transform" is actually a Fourier series: its support is in $\mathbb{Z}$. This follows by noting that
$$\begin{align}C(x) &= \int_{-\infty}^{\infty} \hat{C}(\xi)e^{2\pi i x \xi}d\xi; \\ C(x) &= \sum_{n\in \mathbb{Z}}a_n e^{2\pi i n x}; \end{align}$$
Where the first line is the Fourier inversion formula and the second line is the Fourier series for $C$. It follows by uniqueness that $\hat{C}(\xi) = \sum_{n\in \mathbb{Z}}a_n \delta(\xi - n)$. On the other hand, the (inverse) Fourier transform of $\hat{C}$ is also supported on $\mathbb{Z}$, so $\hat{C}$ is also periodic with period 1. Thus, all the $a_n$ are the same:
$$\hat{C}(\xi) = a\sum_{n\in \mathbb{Z}}\delta(\xi - n),$$ where $a$ is some scalar. It's not hard to see that the scalar has to be 1.
To derive Poisson summation from this, use the convolution theorem: let $f$ be any function. On the one hand, $$(f*C)(x) = \sum_{n\in \mathbb{Z}} f(x+n).$$ On the other hand, we can use the convolution theorem: $$\widehat{(f*C)}(\xi) = \hat{f}(\xi)\hat{C}(\xi) = \hat{f}(\xi)\sum_{n\in \mathbb{Z}} \delta(\xi-n) = \sum_{n\in \mathbb{Z}} \hat{f}(n)\delta(\xi-n).$$ The last sum gives the Fourier series of the periodic function $f*C$: $$(f*C)(x) = \sum_{n\in \mathbb{Z}} \hat{f}(n)e^{2\pi i n x}.$$ Plugging in $x=0$ gives the Poisson summation formula, QED. But the result for general $x$ is interesting as well: given a function $f$, you can obtain a periodic function $g(x)$ by (a) adding up $f(x+n)$ over all integers $n$, or (b) taking the Fourier transform of $f$ at integer frequencies and making that the Fourier series of $g$. The result is that (a) and (b) give the same function.
It is a special case of the trace formula. Both sides are the trace of the same operator.
A variant on @Darsh Ranjan's argument: since $u=\sum_{n\in\mathbb Z}\delta(x-n)$ is invariant under translation by $\mathbb Z$, and is annihilated by multiplication by $e^{2\pi inx}-1$ for all $n\in \mathbb Z$ (this captures the order-zero aspect!) it suffices to show that there is a unique (up to scalar multiples) distribution meeting these conditions, and that the Fourier transform of $u$ also does meet them. The latter is immediate, since Fourier transform interchanges the two conditions.
For uniqueness: annihilation by multiplication by $e^{2\pi ix}-1$ implies that the support is $\mathbb Z$. We know the classification of distributions supported at points, hence, supported on a discrete closed subset: Dirac deltas and derivatives. Using smooth cut-offs to examine the behavior at a given integer, since $e^{2\pi ix}-1$ vanishes just to order $1$, the order of the $n$-th component is $0$. Thus, such a distribution is of the form $\sum_n c_n\cdot \delta(x-n)$. Translation invariance implies that all the coefficients are the same.