Using your expression, i.e., where you multiplied both sides of the expression for $G'_A(x)$ by $x$, you could replace all $x$ with $x^m$ to get
$$\begin{equation}\begin{aligned}
x\left(G'_A(x)\right) & = \sum_{n \epsilon A} f(n) x^n \\
x^m\left(G'_A(x^m)\right) & = \sum_{n \epsilon A} f(n) \left(x^m\right)^n \\
G'_A(x^m)x^m & = \sum_{n \epsilon A} f(n) x^{mn}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Alternatively, you could do the substitution first and then the multiplication, with this, showing it in detail, resulting in
$$\begin{equation}\begin{aligned}
G_A(x) & = \sum_{n\epsilon A} \frac{f(n)}{n} x^{n} \\
G'_A(x) & = \sum_{n\epsilon A} f(n) x^{n-1} \\
G'_A(x^m) & = \sum_{n\epsilon A} f(n) \left(x^m\right)^{n-1} \\
G'_A(x^m) & = \sum_{n\epsilon A} f(n) x^{mn - m} \\
G'_A(x^m)x^m & = \sum_{n\epsilon A} f(n) x^{mn - m}x^m \\
G'_A(x^m)x^m & = \sum_{n\epsilon A} f(n) x^{mn}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
In either case, taking the sum from $m = 1$ to $\infty$ of both sides gives the part you're asking about, i.e,.
$$\sum_{m=1}^{\infty} G'_A(x^m)x^m = \sum_{m=1}^{\infty} \sum_{n\epsilon A} f(n) x^{mn} \tag{3}\label{eq3A}$$
