I think the answer is 'yes.' I don't have a suitably general reason why this is the case, although surely one exists and is in the literature somewhere.
At any rate, for the problem at hand, we have for $s > 0$
$$\sum \frac{a_n}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty \sum a_n e^{-nt} t^{s-1} dt.$$
Edit: the interchange of limit and sum used here requires justification, and this is done below. Supposing that $\sum a_n x^n \rightarrow \sigma$, we may write $\sum a_n e^{-nt} = (\sigma + \epsilon(t))\cdot e^{-t}$ where $\epsilon(t) \rightarrow 0$ as $t \rightarrow 0$, and $\epsilon(t)$ is bounded for all t. In this case
$$\sum \frac{a_n}{n^s} = \sigma + O\left(s\int_0^\infty \epsilon(t) e^{-t} t^{s-1} dt\right)$$
Showing that error term tends to $0$ is just a matter of epsilontics; for any $\epsilon > 0$, there is $\Delta$ so that $|\epsilon(t)| < \epsilon$ for $t < \Delta$. Hence
$$\left|s\int_0^\infty \epsilon(t) e^{-t} t^{s-1} dt \right| < s\epsilon \int_0^\Delta t^{s-1} dt + s\int_\Delta^\infty e^{-t}t^{s-1}dt < \epsilon \Delta^s + s\int_\Delta^\infty e^{-t} t^{-1} dt.$$
Letting $s \rightarrow 0$, our error term is bounded by $\epsilon$, but $\epsilon$ of course is arbitrary.
Edit: Justifying the interchange of limit and sum above is surprisingly difficult. We will require
Lemma: If for fixed $\epsilon > 0$, the partial sums $D_{\epsilon}(N) = \sum_{n=1}^N a_n/n^\epsilon = O(1),$ then
(a) $A(N) = \sum_{n \leq N} a_n = O(n^\epsilon)$, and
(b) $\sum_{n \leq N} a_n e^{-nt} = O(t^{-\epsilon})$,
where the O-constants depend on $\epsilon.$
This, with the hypothesis that $\sum a_n/n^s$ converges for all $s > 0$, imply the conclusions a) and b) for all positive $\epsilon$.
To prove part a), note that
$$\sum_{n \leq N} a_n = \sum_{n \leq N} a_n n^{-\epsilon}n^\epsilon = \sum_{n \leq N-1} D_{\epsilon}(n) (n^\epsilon - (n+1)^\epsilon) + D_\epsilon(N)N^\epsilon,$$
which is seen to be $O(N^\epsilon)$ upon taking absolute values inside the sum.
To prove part b), note that
$$t^\epsilon \sum_{n \leq N} a_n e^{-nt} = t^\epsilon \sum_{n=1}^{N-1} A(n)(e^{-nt} - e^{-(n+1)t}) + t^\epsilon A(N) e^{-Nt} = O \left( \sum_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) + (tN)^\epsilon e^{-Nt}\right).$$
Now, $(tN)^\epsilon e^{-Nt} = O(1)$, and
$$\sum_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) = 2^\epsilon(1-e^{-t}) \sum_{n\leq N} (tn/2)^\epsilon e^{-nt/2} e^{-nt/2} = O\left(\frac{1-e^{-t}}{1-e^{-t/2}}\right) = O\left(\frac{1}{1+e^{t/2}}\right) = O(1),$$
and this proves b).
We use this to justify interchanging sum and integral as follows: note that
$$\sum_{n=1}^N \frac{a_n}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt,$$
and therefore
$$\frac{1}{\Gamma(s)}\int_0^\infty \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt = \frac{1}{\Gamma(s)}\int_0^1 \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt + \frac{1}{\Gamma(s)}\int_1^\infty \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt.$$
In the first integral, note that for $\epsilon < s$, $\sum_{n \leq N} a_n e^{-nt} t^{s-1} = O(t^{s-\epsilon -1})$ for all $N$. So by dominated convergence in the first integral, and uniform convergence of $e^t \sum_{n=1}^N a_n e^{-nt}$ for $t \geq 1$ in the second, this is limit is
$$\lim_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int_0^1 \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt + \lim_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int_1^\infty \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt = \lim_{N\rightarrow\infty} \sum_{n=1}^N a_n \frac{1}{\Gamma(s)}\int_0^\infty e^{-nt}t^{s-1} dt.$$
This is just $\sum_{n=1}^\infty \frac{a_n}{n^s}$.
Note then that we do not need to assume from the start that the infinite Dirichlet sum tends to anything as $s \rightarrow 0$; once it converges for each fixed $s$, that is implied by the behavior of the power series.
