= Solution
Conditioning on the intensity in the <Poisson mixture> gives
$$
G_N(z)=\mathbb E[e^{\Lambda(z-1)}]
=\left(\frac{p}{1-qz}\right)^2.
$$
Thus $N$ has the <negative binomial distribution> with two successes and success probability $p$, counting failures; explicitly $\mathbb P(N=j)=(j+1)p^2q^j$ for $j\geq0$. The <probability generating function> is finite for real $z<1/q$.
The claim-size <moment-generating function> is $M_X(t)=(1-\mu t)^{-1}$. Substituting into the aggregate <moment-generating function> and using $p+q=1$ gives
$$
\boxed{M_S(t)=\left(\frac{p(1-\mu t)}{p-\mu t}\right)^2},
\qquad t<\frac p\mu.
$$
Put $L(t)=p/(p-\mu t)$, the <moment-generating function> of an <exponential distribution> with rate $p/\mu$. The identity
$$
\frac{p(1-\mu t)}{p-\mu t}=p+qL(t)
$$
then yields
$$
M_S(t)=p^2+2pqL(t)+q^2L(t)^2.
$$
The <gamma-mixed Poisson aggregate with exponential claims> has three nonnegative <mixture weights> summing to one. By uniqueness of the <moment-generating function> near zero, \b[the aggregate distribution is]
$$
\boxed{\mathcal L(S)=p^2\delta_0+
2pq\,\operatorname{Exp}(p/\mu)+
q^2\,\operatorname{Gamma}(2,\text{rate }p/\mu).}
$$
Here $\delta_0$ is the <Dirac measure> at zero. In particular $\mathbb P(S=0)=p^2$, consistently with $\mathbb P(N=0)$. The positive components have respective <expected values> $\mu/p$ and $2\mu/p$; their <mixture distribution> accounts for the possibility of no aggregate payout.
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