As long as the singularities are not "too bad", the answer will be yes, $f(x)$ will represent a solution of the differential equation. The very same way that $$ \sum_{x=0}^{\infty} (-1)^{x}n!x^{n+1} $$ represents a solution of $x^2y'+y=x$. One might object that the series diverges, but resummation theory says this is irrelevant, that sum nevertheless represents a unique function (in a sector) which is a solution of the differential equation.

Balser's book, "From divergent series to analytic differential equations", would be a good place to start. Since the theory fundamentally uses Mellin transforms, you should be able to find what you want (perhaps indirectly) there.

As long as the singularities are not "too bad", the answer will be yes, $f(x)$ will represent a solution of the differential equation. The very same way that $$ \sum_{n=0}^{\infty} (-1)^nn!x^{n+1} $$ represents a solution of $x^2y'+y=x$. One might object that the series diverges, but resummation theory says this is irrelevant, that sum nevertheless represents a unique function (in a sector) which is a solution of the differential equation.

Balser's book, "From divergent series to analytic differential equations", would be a good place to start. Since the theory fundamentally uses Mellin transforms, you should be able to find what you want (perhaps indirectly) there.