I asked this question the other day and got a sufficient answer: that given $X$~$\mathsf{Poisson}(rate=\lambda_1)$ and $Y$~$\mathsf{Poisson}(rate=\lambda_2)$, $X|(X+Y=n)$~$\mathsf{Binomial}\left(n,\cfrac{\lambda_1}{\lambda_1+\lambda_2}\right)$. However, the way it was done was not using moment generating functions, and indeed I got a moment generating function for $X|(X+Y=n)$: $e^{tn}\cdot e^{\lambda_2(e^{-t}-1)}$. Knowing that a binomial's MGF looks like $(pe^t+1-p)^n$, I tried to prove the same result using the moment generating function I got, and I got close but got stuck. Here's what I did:
$\left(\cfrac{\lambda_1}{\lambda_1+\lambda_2}e^t+1-\cfrac{\lambda_1}{\lambda_1+\lambda_2}\right)^n \stackrel{?}=e^{tn}\cdot e^{\lambda_2(e^{-t}-1)}\iff n=\mathsf{log}_{\cfrac{\lambda_1}{\lambda_1+\lambda_2}\cdot(e^t-1)+1}(e^{\lambda_2(e^{-t}-1)+tn}) \\ = \cfrac{\mathsf{ln}(e^{\lambda_2(e^{-t}-1)+tn})}{\mathsf{ln}\left(\cfrac{\lambda_1}{\lambda_1+\lambda_2}(e^t-1)+1\right)} = \cfrac{\lambda_2(e^{-t}-1)+tn}{\mathsf{ln}\left(\cfrac{\lambda_1(e^t-1)}{\lambda_1+\lambda_2}+\cfrac{\lambda_1+\lambda_2}{\lambda_1+\lambda_2}\right)} = \cfrac{\lambda_2(e^{-t}-1)+tn}{\mathsf{ln}\left(\cfrac{\lambda_1e^t + \lambda_2}{\lambda_1+\lambda_2}\right)} = \cfrac{\lambda_2(e^{-t}-1)+tn}{\mathsf{ln}(\lambda_1e^t+\lambda_2) - \mathsf{ln}(\lambda_1+\lambda_2)} = \cfrac{\lambda_2(e^{-t}-1)+tn}{\mathsf{ln}(e^t)+\mathsf{ln}(\lambda_1+\lambda_2e^{-t}) - \mathsf{ln}(\lambda_1+\lambda_2)} = \cfrac{\lambda_2(e^{-t}-1)+tn}{t+\mathsf{ln}(\lambda_1+\lambda_2e^{-t}) - \mathsf{ln}(\lambda_1+\lambda_2)}$
Multiplying both sides by the denominator,
$tn+n\cdot\mathsf{ln}(\lambda_1+\lambda_2e^{-t})-n\cdot\mathsf{ln}(\lambda_1+\lambda_2) = \lambda_2(e^{-t}-1)+tn \iff n\cdot\mathsf{ln}(\lambda_1+\lambda_2e^{-t})-n\cdot\mathsf{ln}(\lambda_1+\lambda_2) = \lambda_2(e^{-t}-1)$
This is where I'm stuck. I can't seem to get rid of the $\lambda_1$. Thanks for the help.
You made an unjustified step in you efforts to obtain the MGF of $X\mid Z$.
$\begin{align}\mathsf E(\mathrm e^{tX}\mid X=n-Y) &= \mathsf E(\mathrm e^{t(n-Y)}\mid Y=n-X) \\[1ex]&\neq \mathsf E(\mathrm e^{t(n-Y)})\end{align}$
The very purpose of the exercise suggests that the conditional expectation of the variable $Y$ under the event of $X+Y=n$, may not be Poisson (or at very least not with the same parameter), so there is no justification for presuming that the conditional expectation of a function of $Y$, given that event, should be equal to the unconditional expectation of that function.
Hence $\mathrm e^{nt}\mathrm e^{\lambda_2(\mathrm e^{-t}-1)}$ is in fact not the MGF .