Extracting Dirichlet series coefficients

Yes. If $f(s)$ has a finite abscissa of absolute convergence $\sigma_a$, then $\forall \sigma > \sigma_a$: $$ \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} f(\sigma+ it)n^{it} \mathrm{d}t = \frac{a_n}{n^{\sigma}}. $$ IIRC, the proof can be found in Apostol's book on Analytic Number Theory.

Even for more general Dirichlet series $$f(z)=\sum_{0}^\infty a_n e^{-\lambda_nz}$$ there is the formula $$a_ne^{-\lambda_n\sigma}=\lim_{T\to\infty}\frac{1}{T}\int_{t_0}^Tf(\sigma+it)e^{\lambda_n it}dt,$$ where $t_0$ is arbitrary (real) and $\sigma>\sigma_u$, the abscissa of uniform convergence.

This formula determines both $\lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{i\lambda t}$ with $\lambda\neq \lambda_n$. The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line $\{s=\sigma+it:t\in R\}$). The "number-theoretic case" corresponds to $\lambda_n=n$.

Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969. Dirichlet series with complex $\lambda_n$ have been also studied (by A. F. Leont'ev and his school).