For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have $$ \tilde{f}(s) = (2\pi)^{-s} \Gamma(s) L(s,f). $$ My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.
I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?
If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i \infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n \geq 1$. This implies that $f=g$ (having the same Laurent series).