Let $F(n) = \sum f(n)n^{-s}$ that converges absolutely for $\sigma > \sigma_a$, where $f(1) \not= 0$. We may define the Dirichlet series, $G(s) := \sum f^{-1}(n)n^{-s}$, where $f^{-1}$ is the Dirchlet inverse of $f$, satisfiying $f * f^{-1}(n) = [\frac{1}{n}]$.
I was wondering if $$ \text{ $\sum f^{-1}(n)n^{-s}$ converges absolutely for $\sigma > \sigma_a$ } $$
If you can show that $f^{-1}(n)$ is itself a Dirichlet character, then $L(s,f^{-1})$ becomes a Dirichlet series which will definitely converge for $\Re(s)>1$ (Apostol, eq. (2), p. 224). You may then be able to use something like Theorem 2.8 (Apostol, p.30) and Theorem 11.1 (p.225) to help determine $\sigma_{a'}$ and possibly $a'=a$. But as mentioned in the comments the abscissa of convergence can be different.