Let $f(s)$ be meromorphic in $\mathbb{C}$. Let the following inverse Mellin transform be Lebesgue integrable for all real positive $x$ at some complex point $s$ with some real $c$:
$\frac{1}{2\pi i} \int\limits_{c-i\infty}^{c+i\infty} f(s+w)\frac{x^w}{w}dw$
Does the associated classical Dirichlet series of the form $\sum_{n=1}^\infty a_n n^{-s}$ exist, such that for all real positive $x$
$\sum_{n\leq x} \frac{a_n}{n^s}=\frac{1}{2\pi i} \int\limits_{c-i\infty}^{c+i\infty} f(s+w)\frac{x^w}{w}dw$
If so, under what conditions on $f(s)$? Any book reference?
Edit1: Let $s$ and $w$ be independent complex variables and let the above integral be Lebesgue integrable on open strips $r_1<\Re s < r_2$ and $R_1<\Re w < R_2$.
Your question is unclear. If $w \mapsto \frac{f(s+w)}{w}$ is analytic and $L^1$ on the vertical lines $\Re(w) \in (c-a,c+a)$ then $h(x) = \frac{1}{2\pi i} \int_{-i\infty}^{c+i\infty} \frac{f(s+w)}{w} x^wdw$ is continuous (so it is not the partial sums of a Dirichlet series) and has decay $h(e^u) = O(e^{-(a-\epsilon)|u|})$ and $\frac{f(s+w)}{w} = \int_0^\infty x^{-w-1}h(x)dx$ for $\Re(w) \in (c-a,c+a)$. If $\frac{f(s+w)}{w}$ is only $L^1$ or $L^2$ or $L^\infty$ on one vertical line then the latter Mellin transform exists in the $L^2$ sense or in the sense of distributions.
If $w \mapsto f(s+w)$ is a Dirichlet series which converges absolutely for $\Re(w) > d \ge 0$ then $\frac{f(s+w)}{w}$ is $L^2$ on vertical lines $\Re(w) > d$ and it $\to 0$ uniformly as $\Re(w) \to \infty$. From such conditions on $f$ then $h(x) x^{-d-\epsilon}$ is $L^2$ and supported on $x \ge 1$ and there is no obvious way to check if $h$ is constant on intervals $(n,n+1)$ ie. that $f$ is a Dirichlet series.
Start from any continuous function $h$ decreasing fast enough but not constant on intervals $(n,n+1)$ then $$w\int_0^\infty h(\lfloor x\rfloor)x^{-w-1}dx= \sum_{n=1}^\infty (h(n)-h(n-1))n^{-w}$$ is a Dirichlet series but $w\int_0^\infty h( x)x^{-w-1}dx$ is not. In your post $s$ is constant.
Once $h$ the inverse Mellin transform of $F(w)=f(s+w)/w$ exists then $F,h$ are uniquely determined by each other (up to the problem of domain convergence) and you know $f(w)$ and $f(z+w)/w$ for any $z$.