We know that
$$ f(s)= \int_{-\infty}^{\infty}f(x)\delta (x-s) d x$$
however, is there a similar delta function so for the Mellin transform
$$ f(s)=\int_{0}^{\infty}f(x)m(xs) d x$$ ?
That is a function $ m(x) $ or distribution with similar properties of the delta function but for the Mellin transform.
The expression
$$ f(s) = \int_0^\infty f(x) m(xs) \mathrm{d}x $$
cannot be satisfied for all $f$. Let $g(x) = f(\lambda x)$ for $\lambda > 0$. Then
$$ \int_0^\infty g(x) m(xs) \mathrm{d}x = \int_0^\infty f(\lambda x) m(\lambda x \cdot \frac{s}{\lambda}) \frac{1}{\lambda}\mathrm{d}(\lambda x) = \frac{1}\lambda\int_0^\infty f(y) m(y s/\lambda) \mathrm{d}y = \frac{1}{\lambda} f(s/\lambda) \neq g(s) = f(s\lambda)$$
in general.