Given a function $f: \mathbb{R}^+ \rightarrow \mathbb{C}$ satisfying suitable conditions (exponential decay at infinity, continuous, and bounded variation) is good enough, its Mellin transform is defined by the function

$$M(f)(s) = \int_0^{\infty} f(y) y^s \frac{dy}{y},$$

and $f(y)$ can be recovered by the Mellin inversion formula:

$$f(y) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} y^{-s} M(f)(s) ds.$$

This is a change of variable from the Fourier inversion formula, or the Laplace inversion formula, and can be proved in the same way. This is used all the time in analytic number theory (as well as many other subjects, I understand) -- for example, if $f(y)$ is the characteristic function of $[0, 1]$ then its Mellin transform is $1/s$, and one recovers the fact (Perron's formula) that

$$\frac{1}{2\pi i} \int_{2 - i \infty}^{2 + i \infty} n^{-s} \frac{ds}{s}$$

is equal to 1 if $0 < n < 1$, and is 0 if $n > 1$. (Note that there are technical issues which I am glossing over; one integrates over any vertical line with $\sigma > 0$, and the integral is equal to $1/2$ if $n = 1$.)

I use these formulas frequently, but... I find myself having to look them up repeatedly, and I'd like to understand them more intuitively. Perron's formula can be proved using Cauchy's residue formula (shift the contour to $- \infty$ or $+ \infty$ depending on whether $n > 1$), but this proof doesn't prove the general Mellin inversion formula.

My question is:

What do the Mellin transform and the inversion formula mean? Morally, why are they true?

For example, why is the Mellin transform an integral over the positive reals, while the inverse transform is an integral over the complex plane?

I found some resources -- Wikipedia; this MO question is closely related, and the first video in particular is nice; and a proof is outlined in Iwaniec and Kowalski -- but I feel that there should be a more intuitive explanation than any I have come up with so far.