= Solution
Let $\Lambda=\sum_{i=1}^m\lambda_i$. For $\Lambda>0$, the resulting <compound Poisson distribution> has
$$
\boxed{\text{count parameter }\Lambda,\qquad
F(x)=\sum_{i=1}^m\frac{\lambda_i}{\Lambda}F_i(x).}
$$
Thus the merged <claim size> <probability distribution> is a rate-weighted <mixture distribution>, not an equally weighted <mixture distribution> unless the rates agree.
To prove this without any <moment-generating function> assumption, let $\chi_i(t)$ be the <characteristic function> of a claim of type $i$. Conditioning on its <Poisson distribution> count gives
$$
\mathbb Ee^{itS_i}=\exp\{\lambda_i(\chi_i(t)-1)\}.
$$
<Independence> of the aggregate risks therefore gives
$$
\mathbb Ee^{itS}
=\prod_i\exp\{\lambda_i(\chi_i(t)-1)\}
=\exp\left\{\Lambda\left(\sum_i\frac{\lambda_i}{\Lambda}\chi_i(t)-1\right)\right\}.
$$
The weighted <characteristic function> inside is precisely that of the displayed <mixture distribution>. This is the transform of a sum of <independent> claims with <independent> count $\operatorname{Pois}(\Lambda)$, proving the claim by uniqueness of <characteristic functions>. This is <Poisson superposition of insurance portfolios>. If $\Lambda=0$, all risks make no claims <almost surely> and $S=0$; the severity law is then irrelevant.
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