I have $X,Y\sim Poisson(\lambda)$ and if I observe $X=0$ first, no $Y$ is observed. If $X>0$, then we observe a $Y$.
I want the pmf of $T=0$ if $X=0$ and $T=1+X+Y$ if $X>0$.
I know the joint mass function of $X,Y$ is $$\displaystyle\frac{\lambda^x e^{-\lambda }}{x!}\left(\frac{\lambda ^y e^{-\lambda }}{y!} \right) ^{\{x>0\}}$$
Then $P(T=0)=P(X=0)=e^{-\lambda}$
but if $X>0$,
$P(T=t)=P(1+X+Y=t)=P(X+Y=t-1)=P(X=x, Y=t-x-1)=\displaystyle\frac{\lambda^{t-1}e^{-2\lambda}}{x!(t-x-1)!}$
Although in general correct, in the last equation you need to sum over the possible $x$'s, that is $$P(T=t)=\ldots=P(X+Y=t-1)=\sum_{x=1}^{t-1}P(X=x,Y=t-x-1)=\sum_{x=1}^{t-1}\frac{\lambda^{t-1}e^{-2\lambda}}{x!(t-x-1)!}$$ General note: This would be correct if we assume that $T=1+X+Y$ (as you have done), but as you state the problem I would go for $T=X+Y$. I do not understand where this $1$ comes from.