I haven't checked this, so I have probably gotten some of the regions of validity wrong, but here are my thoughts on approximating the sum.
Write $N = \frac{\log a}{\log x}$ so that your series becomes $\sum_{n=0}^\infty \frac{x^n}{a+x^n} = \sum_{n=0}^\infty \frac{1}{1+x^{N-n}}$. Over all integer $n$, this curve is a sigmoid, and $N$ is the knee of the curve (value $1/2$) where terms transition from "near 1" to "near 0"; you can approximate or bound the series separately in these two regimes.
Things are especially convenient if you can restrict $2N$ to an integer; $\frac{1}{1+x^{N-n}}+\frac{1}{1+x^{N-(2N-n)}}=1$, so terms $n$ and $2N-n$ of the series now sum to $1$ and the sum from $0$ to $2N$ is $\frac{2N+1}{2}$ exactly, and you only have to consider the tail of the sum. But even if $2N$ is not an integer, this is a pretty good approximation for the first $\round{2N}$ terms of the sum (including a fraction of the final term).
If $1/a$ or $1/x$ is large, then in the tail of the sum ($n>2N$) you can reasonably approximate $\frac{1}{1+x^{N-n}} \approx x^{n-N}-x^{2(n-N)}+\cdots$ using the first few terms of a geometric series; now just sum these to get an estimate of the tail of the series. This estimate isn't very good when $a$ and $x$ are near $1$; but for $x$ near $1$ the terms vary slowly with $n$, and you can approximate the sum by an integral (which evaluates to a hypergeometric function).
