= Solution
For <excess of loss reinsurance> with retention $M$, the direct insurer caps its payment at $M$ and the reinsurer pays the excess. Thus
$$
\boxed{Y_i=\min(X_i,M),\qquad Z_i=(X_i-M)_+.}
$$
Let $I_i=\mathbf1_{\{X_i>M\}}$ and $\alpha=\mathbb P(X_i>M)=1-F_{X_1}(M)\in(0,1)$. These are independent <Bernoulli random variables>, independent of $N$. Since $N_R=\sum_{i=1}^NI_i$, its conditional law given $N=n$ is a <binomial distribution>. Conditioning on $N$ therefore gives the <Bernoulli thinning> identity
$$
\mathbb E[z^{N_R}\mid N]=(1-\alpha+\alpha z)^N,\qquad\boxed{G_{N_R}(z)=G_N(1-\alpha+\alpha z).}
$$
Here the <probability generating function> may be evaluated on $0\le z\le1$, and the identity extends wherever the transforms converge.
The positive reinsurer marks have the conditional law $W\overset d=X_1-M\mid X_1>M$. Hence
$$
\boxed{F_W(w)=\begin{cases}0,&w<0,\\\dfrac{F_{X_1}(M+w)-F_{X_1}(M)}{1-F_{X_1}(M)},&w\ge0,\end{cases}\qquad f_W(w)=\frac{f_{X_1}(M+w)}{1-F_{X_1}(M)}\quad(w>0).}
$$
The independent-mark representation is justified as follows. Conditional on any fixed count and pattern of exceedance indicators, the positive excesses are independent with this same conditional law; that law depends neither on the number of exceedances nor on their locations. Averaging over the indicator pattern thus leaves the positive mark sequence independent of $N_R$. We may extend it by additional independent copies if needed, giving $S_R=\sum_{j=1}^{N_R}W_j$ with the stated independence.
Now write $q=1-p$. For <exponential distribution> claim sizes of mean $\mu$,
$$
\alpha=e^{-M/\mu},\qquad\mathbb P(W>w)=\frac{e^{-(M+w)/\mu}}{e^{-M/\mu}}=e^{-w/\mu}.
$$
Thus \b[$W$ is exponential with the original mean $\mu$], by <memorylessness of the exponential distribution>. The original zero-based <geometric distribution> count has $G_N(z)=p/(1-qz)$. Consequently
$$
G_{N_R}(z)=\frac{p}{p+q\alpha-q\alpha z}=\frac{p_R}{1-q_Rz},\qquad p_R=\frac{p}{p+q\alpha},\quad q_R=\frac{q\alpha}{p+q\alpha}.
$$
This is the <geometric count under Bernoulli thinning>, with
$$
\boxed{\mathbb P(N_R=k)=p_Rq_R^k\quad(k=0,1,\ldots).}
$$
Finally use the <Laplace transform of a nonnegative random variable>, keeping the mass at zero. For $t\ge0$,
$$
\mathbb E e^{-tS_R}=G_{N_R}\!\left(\frac1{1+\mu t}\right)=\frac{p_R(1+\mu t)}{p_R+\mu t}=p_R+q_R\frac{p_R}{p_R+\mu t}.
$$
The last expression is the transform of a <mixture distribution>: mass $p_R$ at zero and, with probability $q_R$, an <exponential distribution> of rate $p_R/\mu$. Uniqueness of the <Laplace transform> identifies this <zero-based geometric sum of exponential variables>:
$$
\boxed{\mathcal L(S_R)=p_R\delta_0+q_R\operatorname{Exp}(p_R/\mu).}
$$
Equivalently, $\mathbb P(S_R=0)=p_R$ and its positive density is $q_R(p_R/\mu)e^{-p_Rs/\mu}$. Therefore \b[the requested exceedance probability] is
$$
\boxed{\mathbb P(S_R>s)=\frac{(1-p)e^{-M/\mu}}{p+(1-p)e^{-M/\mu}}\exp\!\left(-\frac{p\,s}{\mu[p+(1-p)e^{-M/\mu}]}\right),\qquad s>0.}
$$
The prefactor is essential: the aggregate is zero whenever no claim crosses the retention.
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