This is an approach that I found by "googling", "papering" and "wikipeding". When I realized the idea, I imagine myself as the first mathematicians looking at the way in which physicists made use of dirac distribution (at the very beginning).

What I mean, is that I'm still not able to justify any step of what follows, but according to many papers in Quantum Field Theory (a theory of which I do not know anything) it works in many many situations.

The idea is the following one (and believe me, to reach the following "cleaner" description was a terrible task).

Let us soppose that $\lim_{x\to\infty} S(x)=\ell\in
\mathbb{R}-\{0\}$. This means that
\begin{equation}
\frac{1}{S(x)}=\frac{x}{x\sum_{n=0}^{\infty} a_n x^n}=\frac{x}{\sum_{n=1}^{\infty} b_n x^n}\to\frac{1}{\ell} \quad \text{when } x\to+\infty .
\end{equation}
Here we have defined $b_n:=a_{n-1}$ for every $n\in\mathbb{N}$, $n\geq 1$.
In particular, for every $M\in\mathbb{N}$ one has
\begin{equation}
\frac{1}{S(x)}=\left(\frac{x^M}{\sum_{n=1}^{\infty} c(M)_n x^n}\right)^{1/M}
\end{equation}
where the sequence $(c(M))_n$ is given by the "M-times" *polynomial* convolution of the sequence $(b_n)$. Therefore
\begin{equation}
\frac{1}{S(x)}=\lim_{M\to\infty}\left(\frac{x^M}{\sum_{n=1}^{\infty} c(M)_n x^n}\right)^{1/M} .
\end{equation}

Now starts the "strange" part. One writes
\begin{equation}
\frac{1}{S(x)}=\lim_{M\to\infty}\lim_{N\to\infty}\left(\frac{x^M}{\sum_{n=1}^{N} c(M)_n x^n}\right)^{1/M}.
\end{equation}
Next one choose $M=N$ to obtain
\begin{equation}
\frac{1}{S(x)}=\lim_{N\to\infty}\left(\frac{x^N}{\sum_{n=1}^{N} c(N)_n x^n}\right)^{1/N}.
\end{equation}
To finish, one pass to the limit for $x\to\infty$, and interchange the limits:
\begin{equation}
\lim_{x\to+\infty}\frac{1}{S(x)}=\lim_{N\to\infty}\left(\lim_{x\to+\infty}\frac{x^N}{\sum_{n=1}^{N} c(N)_n x^n}\right)^{1/N}=\lim_{N\to\infty} \frac{1}{c(N)_N^{1/N}} .
\end{equation}
Thus
\begin{equation}
\ell=\lim_{N\to\infty}c(N)_N^{1/N}.
\end{equation}

1more comment