this is an (incomplete) sketch of proof I planned to use at first for answering to The value of a Dirichlet L-function at s=0.
let $F(s) = \sum_{n=1}^\infty a_n n^{-s}$.
- suppose it is converging for $Re(s) > 0$ and $F(0) = \lim_{s \to 0^+}F(s)$ exists. $a_n = \mathcal{O}(1)$ and $f(x) = \sum_{k=1}^\infty a_n x^n$ converges for $|x| < 1$.
- suppose also that $f$ doesn't have a singularity at $s= 1$.
from $n^{-s} \Gamma(s) = n^{-s} \int_0^\infty y^{s-1} e^{-y} dy = \int_0^\infty x^{s-1} e^{-nx}dx$ (with the change of variable $y = nx$) and because when $s \to 0^+$ : $\Gamma(s) \sim \frac{1}{s}$ we get when $s \to 0^+$:
$$\frac{F(0)}{s} \sim \Gamma(s) F(s) = \sum_{n=1}^\infty a_n n^{-s} \Gamma(s) = \sum_{n=1}^\infty a_n \int_0^\infty x^{s-1} e^{-nx} dx = \int_0^\infty x^{s-1} f(e^{-x})dx\qquad(1)$$
(inverting $\int$ and $\sum$ by dominated convergence). since $|f(e^{-x})|$ is bounded, that last integral converges for $Re(s) > 0$, and for any $\epsilon > 0$ : $\int_\epsilon^\infty x^{s-1} f(e^{-x})dx$ converges when $s = 0$. hence when $s\to 0^+$, using the continuity of $f(x)$ on $[0,1]$ :
$$\int_0^\infty x^{s-1} f(e^{-x}) dx \sim \int_0^1 x^{s-1} f(e^{-x}) dx \sim f(e^{0}) \int_0^1 x^{s-1} dx \sim \frac{f(1)}{s} \qquad(2)$$
hence :
$$(1) \text{ and } (2)\quad\implies \quad F(0) = f(1)$$
from this you should be able to see how the constraints $a_n = \mathcal{O}(1)$ and both $F(0)$ and $f(1)$ exist can be discarded or not.
