The more general question will be what is the criteria for a function $f(t)$ to be the Laplace transform of some random variable $\xi$, i.e. $\mathbb{E}e^{-t\xi}=f(t)$?
A more specific question is why the following is true:
Given a random variable $N\in\mathbb{N}$, $\mathbb{E}N<\infty$, and a sequence of random variables $\xi_i$(independent of $N$) such that $\mathbb{E}\left[\sum_{i\le N}e^{-t^*\xi_i}\right]<\infty$ for some $t^*>0$. Then there is some random variable $\xi$ with Laplace transform $$\mathbb{E}e^{-t\xi} = \frac{\mathbb{E}\left[\sum_{i\le N}e^{-t\xi_i}\right]}{\mathbb{E}N}, t\ge 0.$$
For the specific question, I tried the inverse Laplace transform, but I am not getting anything useful unless $\xi_i$ have the same distribution(then $\mathbb{E}N$ term cancels).
A careless reading of the question and the ambiguity of the term "Laplace transform" led me to propose an answer based on Bernstein's theorem. I hope this is a better answer.
Here is an answer to the specific question. Let $M$ be a random variable independent of the $\xi_i$ such that $P(M=m) = m P(N=m)/E[N]$. Let $I$ be chosen uniformly from $\{1,2,\dots,M\}$, and let $\xi = \xi_I$. The claim is, this $\xi$ has the desired Laplace transform. We have $$ \begin{align} E\exp(-t\xi)&=\sum_{m=1}^\infty P(M=m) E[\exp(-t\xi)|M=m]\\ &= \sum_{m=1}^\infty \frac{m P(N=m)}{E[N]}E[\exp(-t\xi)|M=m]\\ &= \sum_{m=1}^\infty \frac{m P(N=m)}{E[N]} \frac{\sum_{i\le m}E[\exp(-t\xi_i)|N=m]}m\\ &= \frac{\sum_{n=1}^\infty P(N=n)\sum_{i\le n}E[\exp(-t\xi_i)|N=n]}{E[N]}\\ &= \frac{E[\sum_{i\le N}\exp(-t\xi_i)]}{E[N]}, \end{align} $$ which is what is wanted.
Here the distribution of $M$ is a "tilt" of the distribution of $N$, and the independence of $M$ (and $N$) from the $\xi_i$ makes conditioning on the event $[M=m]$ the same as conditioning on the event $[N=m]$, as far as integrals involving the $\xi_i$ is concerned.