= Solution
Condition on the <claim count>, defining the empty sum to be zero. <Independence> gives $\mathbb E[S\mid N=k]=k\mathbb E X_1$. By the <law of total expectation>,
$$
\boxed{\mathbb ES=\mathbb EN\,\mathbb EX_1.}
$$
For nonnegative claims this also follows from <Tonelli theorem>; for signed claims finite $\mathbb EN$ and $\mathbb E|X_1|$ ensure <integrability>. <Conditional independence> likewise gives
$$
\mathbb E[e^{tS}\mid N=k]=\prod_{j=1}^k\mathbb E[e^{tX_j}]=M_{X_1}(t)^k,
\qquad
\boxed{M_S(t)=G_N(M_{X_1}(t)).}
$$
This <random-sum transform identity> holds wherever the composed series is finite, or as an extended nonnegative <expectation>. It is not permission to use an algebraic continuation outside the convergence domain.
For the zero-based <geometric distribution>, summing the <geometric series> gives
$$
G_N(z)=\frac p{1-qz},\qquad \mathbb EN=G_N'(1)=\frac qp.
$$
The <exponential distribution> of mean $\mu$ has $M_{X_1}(t)=(1-\mu t)^{-1}$ for $t<1/\mu$. Therefore
$$
\boxed{\mathbb ES=\frac{q\mu}p,\qquad
M_S(t)=\frac p{1-q/(1-\mu t)}
=\frac{p(1-\mu t)}{p-\mu t}\quad(t<p/\mu).}
$$
The stricter bound $t<p/\mu$ is necessary: the <geometric series> requires $qM_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the <moment-generating function> only on its stated domain.
Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb P(S=0)=p$. Splitting the transform gives
$$
M_S(t)=p+q\frac{p/\mu}{p/\mu-t}.
$$
Uniqueness of the <moment-generating function> identifies a <mixture distribution> with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,
$$
\boxed{\mathbb P(S=0)=p,\qquad f_S(x)=\frac{qp}{\mu}e^{-px/\mu}\quad(x>0).}
$$
The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\geq1$, the sum has <Erlang distribution> density $x^{k-1}e^{-x/\mu}/[\mu^k(k-1)!]$. Multiplying by $pq^k$ and summing over $k$ gives $(pq/\mu)e^{-x/\mu}e^{qx/\mu}$, the same density. This is the <zero-based geometric sum of exponential variables>.
Under <excess of loss reinsurance>, the ceded amount for each claim is $(X_j-M)_+$, and the monthly reduction is their <random sum>. The <tail integral formula for moments> gives
$$
\mathbb E(X_1-M)_+=\int_M^\infty\mathbb P(X_1>x)\,dx
=\mu e^{-M/\mu}.
$$
Conditioning on $N$ again yields
$$
\boxed{\mathbb E\sum_{j=1}^N(X_j-M)_+=\frac{q\mu}p e^{-M/\mu}.}
$$
The retained payment is $\sum_{j=1}^N\min(X_j,M)$; it is generally not $\min(S,M)$.
Under <aggregate stop loss reinsurance>, the reduction is $(S-\widetilde M)_+$. The aggregate tail is $\mathbb P(S>x)=q e^{-px/\mu}$ for $x\geq0$, so
$$
\mathbb E(S-\widetilde M)_+=\int_{\widetilde M}^\infty q e^{-px/\mu}\,dx
=\frac{q\mu}p e^{-p\widetilde M/\mu}.
$$
The prefactor is positive. Equality with the per-claim recovery is equivalent to $p\widetilde M=M$, giving the unique <reinsurance retention>
$$
\boxed{\widetilde M=\frac Mp.}
$$
This is the relation for <equal expected recoveries from excess of loss and stop loss>; since $p<1$, the aggregate <reinsurance retention> exceeds the individual <reinsurance retention>. These are reductions in claim payouts, before any reinsurance premium is deducted.
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