= Solution
Use the measurable partition $B_n=\{N=n\}$ for $n\geq0$, omitting any null cells. For $0\leq s\leq1$, the <random variable> $s^X$ is bounded, so the <law of total expectation> applies without moment assumptions. Conditional on $N=n$, the sum contains exactly $n$ terms. Independence of $N$ from the whole sequence preserves their joint law under this conditioning; independence of the $Y_k$ then gives
$$
\mathbb E(s^X\mid N=n)=\mathbb E\left(\prod_{k=1}^n s^{Y_k}\right)
=\prod_{k=1}^n\mathbb E(s^{Y_k})=[G_Y(s)]^n.
$$
For $n=0$ the <empty sum> is zero and the <empty product> is one, in agreement with the formula. Therefore the <probability generating function of an independent random sum> is
$$
\boxed{G_X(s)=\sum_{n=0}^\infty\mathbb P(N=n)[G_Y(s)]^n=G_N(G_Y(s)).}
$$
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