I was inspired by this question: Poisson distribution with exponential parameter
That is, suppose $N\sim\mathsf{Pois}(\lambda)$ and conditioned on $\{N=n\}$, $X$ has exponential distribution with mean $2n$. What is the distribution of $X$?
The first concern is the case where $N=0$. Since $\mathbb P(N=0) = e^{-\lambda}$, I would conclude that this too is the probability that $X=0$ (so $X$ is not a continuous random variable).
For $N>0$ I applied the law of total probability, i.e. for $t>0$: \begin{align} \mathbb P(X\leqslant t, N>0) &= \sum_{n=1}^\infty \mathbb P(X\leqslant t, N>0\mid N=n)\mathbb P(N=n)\\ &=\sum_{n=1}^\infty \left(\int_0^t \frac1{2n}e^{-\frac1{2n}s}\ \mathsf ds\right)\left(e^{-\lambda}\frac{\lambda^n}{n!} \right), \end{align} for which the best closed form I could find was $$ \sum _{n=1}^{\infty } \frac{e^{-\lambda } \lambda ^n \left(1-e^{-\frac{t}{2 n}}\right)}{n!}. $$
Splitting into two terms, the first is $1-e^{-\lambda}$. The second is inscrutable to me, and expanding the $e^{-\frac t{2n}}$ into a series just makes things uglier. Is there an error in my computations, or a more elegant approach to take here? It seems that interchanging the integral with the sum should be justified (Fubini/Tonelli) here, but I am hesitant to do more computation without ensuring that I haven't overlooked something obvious.
To me, your calculation looks fine, and I do not see how to simplify the expression but if you assume the exponential distribution has mean $\frac{1}{2n}$ then simplification can be done as follows: fix a $t\geq 0$ $$ \mathbb{P}(X> t) = \sum_{n=1}^{\infty}\mathbb{P}(X> t|N=n)\mathbb{P}(N=n) = \sum_{n=1}^{\infty} \left(\int_t^{\infty} {2n}e^{-{2n}s}\ \mathsf ds\right)\left(e^{-\lambda}\frac{\lambda^n}{n!} \right) \\ = \sum_{n=1}^{\infty} e^{-{2n}t} e^{-\lambda}\frac{\lambda^n}{n!} = e^{-\lambda}e^{\lambda e^{-2t}}\sum_{n=1}^{\infty} \left(\lambda e^{-2t}\right)^n \frac{e^{{-\lambda e^{-2t}}}}{n!}=e^{-\lambda}e^{\lambda e^{-2t}}\left(1-e^{{-\lambda e^{-2t}}}\right). $$ In the first equality, I have used your observation that for $\mathbb{P}(X> t|N=0)=0$. If this does not help you let me know I will delete the post.