The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation of the Dirac delta function. See my short note on the Inverse Mellin Transform and the Dirac Delta Function. See also some applications in Dirac's Delta Function and Riemann's Jump Function J(x) for the Primes and The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.
Edwards in Riemann's Zeta Function in Ch. 10 Fourier Analysis Sec. 10.1 Invariant Operators on R+ and Their Transforms gives a nice, more group-theoretic intro to the Mellin transform in line with the other comments in this stream.
Note that whereas $e^{sz}$ is an eigenfunction for $d/dz$ and so the Laplace/Fourier transforms are appropriate for devising an operator calculus for $f(d/dz)$, $z^s$ is an eigenfunction of $zd/dz$ and so the Mellin transform is more appropriate for $f(zd/dz)$ .
Two equations which encapsulate the properties of the Fourier and Mellin transforms:
$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$
$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$.