I'm not sure what you mean by "exponentially distributed with parameter 0" (to me that sounds like a uniform distribution maybe?). Let's just solve the more general case instead.
$$ P(x|A) = \frac{1}{x!}A^{x}e^{-A} $$ $$ P(y|A) = \frac{1}{y!}A^{y}e^{-A} $$ $$ P(A) = \lambda e^{-\lambda A} $$
where $x$ and $y$ are positive integers and $A$ and $\lambda$ are real. Let's define the variable $z=x+y$. We look for $P(x|z=n)$. Using Bayes theorem,
$$ P(x|z=n,A) = \frac{P(z=n|x,A)P(x|A)}{P(z=n|A)} $$
Let's find what the different terms are
$$1) \ \ \ P(z=n|x,A) = \sum_{y=0}^{\infty} P(z=n|x,y,A)P(y|A) = \sum_{y=0}^{\infty} \delta((x+y)-n)P(y|A) = P(y=n-x|A) = \frac{1}{n-x!}A^{n-x}e^{-A}, \ n \ge x$$
$$2) \ \ \ P(z=n|A) = \sum_{x=0}^{\infty}\sum_{y=0}^{\infty} P(z=n|x,y,A)P(y|x,A)P(x|A) = \sum_{x=0}^{\infty}\sum_{y=0}^{\infty} \delta((x+y)-n)P(y|A)P(x|A) = \sum_{y=0}^{n}P(y|A)P(x=n-y|A) = \sum_{y=0}^{n} \frac{1}{y!}A^{y}e^{-A} \frac{1}{n-y!}A^{n-y}e^{-A} = e^{-2A}A^{n}\sum_{y=0}^{n} \frac{1}{n-y!}\frac{1}{y!}$$
Note the change of limits in the sum over $y$. This is because the $\delta$ has a non-zero value only if $n-y \ge 0$. Putting all together we get
$$ P(x|z=n,A) = \frac{\frac{1}{n-x!}\frac{1}{x!}}{\sum_{y=0}^{n} \frac{1}{n-y!}\frac{1}{y!}} , \ n \ge x$$
Now we must take into account that $A$ is also a random variable. However, the result we found does actually not depend on $A$, so
$$ P(x|z=n) = \int_0^\infty P(x|z=n,A)P(A)dA = \frac{\frac{1}{n-x!}\frac{1}{x!}}{\sum_{y=0}^{n} \frac{1}{n-y!}\frac{1}{y!}}, \ n \ge x$$