Let $\sum_{n=1}^{\infty}a_nx^n = f(x)$ and $\sum_{n=1}^{\infty}b_nx^n = g(x)$, both series converging for $|x| < r$. We then have: $$\sum_{n=1}^{\infty}b_nf(x^n) = \sum_{n=1}^{\infty}a_ng(x^n)$$ For what values of $x$ does this hold? Clearly since each $f(x^n)$ or $g(x^n)$ must be defined, we must have $|x^n| < r$ for every $n \ge 1$. In particular $|x| \le 1$.
Now by expanding each side one can easily see that each is a rearrangement of the other. But how can I prove that the first series in absolutely convergent? For in that case the equality follows. I can prove this if at least one of $\sum a_i$ or $\sum b_i$ is absolutely convergent: let the second be such, so by expanding: $$|b_1\sum_{m=1}^{\infty}a_mx^m|+|b_2\sum_{m=1}^{\infty}a_mx^{2m}|+\cdots \le s(|b_1|+|b_2|+\cdots)$$ where $s = \sum_{n=1}^{\infty}|a_n||x|^n$. Now the problem is that $r$ need not be $> 1$ and thus $\sum |a_i|$ or $\sum |b_i|$ may diverge. I tried then to consider the following array: $$ \begin{matrix} |b_1a_1x| & |b_1a_2x^2| & |b_1a_3x^3| & |b_1a_4x^4| & \cdots \\ 0 & |b_2a_1x^2| & 0 & |b_2a_2x^4| & \cdots \\ 0 & 0 & |b_3a_1x^3| & 0 & \cdots \\ \cdots\\ \end{matrix} $$ and so on, which is just the expansion of each sum with its terms arranged in the same row. From the theory of double series, it suffices to prove, for example, that the sum of the columns is convergent or that: $$\sum_{n=1}^{\infty}\left(\sum_{i|n}|a_i||b_{n/i}|\right)x^n < +\infty$$But this seems much harder and I don't even know if it's true in the general case.
For $m, n \ge 1$, $(m-1)(n-1) \ge 0$ so $mn \ge m+n-1$, so if $ 0 < |x| \le 1$ we have $|x|^{mn} \le |x|^{m+n-1}$ and
$$\sum_{n=1}^\infty |b_n| |f(x^n)| \le \sum_{n=1}^\infty \sum_{m=1}^\infty |b_n| |a_m| |x|^{mn} \le |x|^{-1} \sum_{n=1}^\infty |b_n| |x|^n \sum_{m=1}^\infty |a_m| |x|^m$$ Thus if the series for $f(x)$ and $g(x)$ converge absolutely for $|x| < 1$, so does $\sum_{n=1}^\infty b_n f(x^n)$ (and similarly $\sum_{n=1}^\infty a_n g(x^n)$).