I read the statement that, for a multiplicative function $h(n)$ we have $$\sum_{n<x} h(n)n^a \sim R \frac{x^{a+1}}{a+1}$$ where $$R := \prod_p \left(1 - \frac 1p \right) \left( 1 + \frac{h(p)}{p} + \frac{h(p^2)}{p^2} + \cdots \right).$$
Where does it come from and under which conditions on $h$? Is there a way to see the result in "integration" terms? (there is a residue/arithmetic part, but the remainder of the result looks like direct integration)
It is not true in general, try with $h(n)=n^2$ (which gives $R=\infty$), $h(n) = n^{-1/2}$ (which gives $R=0$)
and $\sum_n h(n)n^{-s} =\zeta(s)\zeta(s+i)$ which gives $R=\zeta(1+i)$ (yes $\prod_p \frac1{1-p^{-s}}$ converges for $\Re(s)=1,s\ne 1$ by the PNT because $\sum_p p^{-s}$ looks like $\sum_n \frac{n^{-s}}{\log n}$ and $\sum_{n> x} \frac{n^{-s}}{\log n}$ looks like $\frac{x^{1-s}}{(1-s)\log x}$). If we had $\sum_{n\le x}h(n) = Rx+ o(x)$ then $\sum_n (h(n)-1)n^{-s}$ would be $o(1/\Re(s-1))$ for $\Re(s)>1$ which is absurd. In this example $\sum_{n\le x} h(n)= \zeta(s+i) x+\frac{x^{1+i}}{1+i}+o(x)$.
The conditions are those of the Tauberian theorems proving the PNT (often satisfied for the functions appearing naturally in analytic number theory).
Under much stronger assumptions there is a one line proof, see How to determine the residue of an (arithmetic) Dirichlet series. For $a> -1$ the asymptotic for $\sum_{n\le x} h(n)$ gives the asymptotic for $\sum_{n\le x} h(n) n^a$ by partial summation.