= Solution
Write $P(z)=\mathbb E[z^N]$, $F(z)=\mathbb E[z^{X_1}]$ and $G(z)=\mathbb E[z^{S_N}]$ for the three <probability generating functions>. Since both the claim count and every claim size are positive, $P(0)=F(0)=G(0)=0$. Conditioning on the count and using <independence> gives the <random-sum transform identity>
$$
G(z)=\sum_{n\ge1}p_n F(z)^n=P(F(z)).
$$
The crucial point is that the count recurrence begins at $n=2$. Multiplying it by $n z^{n-1}$ and summing, we obtain
$$
\begin{aligned}
P'(z)-p_1
&=\sum_{n\ge2}(an+b)p_{n-1}z^{n-1}\\
&=\sum_{m\ge1}(a(m+1)+b)p_mz^m\\
&=azP'(z)+(a+b)P(z).
\end{aligned}
$$
Consequently $(1-az)P'(z)=(a+b)P(z)+p_1$. The <chain rule> applied to the aggregate <probability generating function> therefore gives
$$
(1-aF(z))G'(z)=\big((a+b)G(z)+p_1\big)F'(z).
$$
These identities hold inside the unit disk, or as identities of <formal power series>. Since $F(0)=0$, every coefficient of the composition depends on only finitely many count probabilities.
Compare the coefficient of $z^{k-1}$. The product $FG'$ contributes $\sum_{j=1}^{k-1}(k-j)f_jg_{k-j}$, and $F'G$ contributes $\sum_{j=1}^{k-1}j f_jg_{k-j}$. Hence
$$
k g_k=p_1 k f_k+\sum_{j=1}^{k-1}\big(a(k-j)+(a+b)j\big)f_jg_{k-j}.
$$
The required <aggregate recursion for zero-truncated Panjer counts> is \b[initialized by $g_0=0$ and $g_1=p_1f_1$], and for every $k\ge1$ it is
$$
\boxed{g_k=p_1f_k+\sum_{j=1}^{k-1}\left(a+\frac{bj}{k}\right)f_jg_{k-j}.}
$$
The empty sum at $k=1$ is zero. Every term on the right is known or has a smaller aggregate index. Omitting $p_1f_k$ would incorrectly apply the usual <Panjer recursion> with a zero initial value, producing zero for every aggregate probability.
For the <zero-truncated Poisson distribution>, $p_n/p_{n-1}=\lambda/n$ for $n\ge2$, so $a=0$, $b=\lambda$ and $p_1=\lambda/(e^\lambda-1)$, with $\lambda>0$. Thus
$$
\boxed{g_k=\frac{\lambda}{e^\lambda-1}f_k+\frac{\lambda}{k}\sum_{j=1}^{k-1}j f_jg_{k-j},\qquad k\ge1.}
$$
As a direct <probability generating function> check, $P(z)=(e^{\lambda z}-1)/(e^\lambda-1)$ and therefore $G(z)=(e^{\lambda F(z)}-1)/(e^\lambda-1)$; differentiation reproduces this recursion.
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