Say we have $X$~Poisson($\lambda_1$), $Y$~Poisson($\lambda_2$) and $S=X+Y$.
We therefore know $S$~$Poisson(\lambda_1 + \lambda_2)$.
I understand how to calculate E(X|S=s) when S is realised to a discrete value. How would we calculate E(X|S) if S remains a random variable?
I know that the result must be a function of S, but following what we would do if S=s:
E(X|S) = $\sum xP(X=x|S)=\sum x\frac{P(X=x,S)}{P(S)}$
the notation P(S) doesn't make sense. How does one calculate this?
$\mathbb E [X \vert S]$ is a random variable. There are many definitions of this that you can find, but they typically require background in measure-theoretic probability in order to deal with cases more general than the one you present.
Here is a simple definition that works here. First, define the function
$$ g(s) = \mathbb E [X \vert S=s].$$
Then we define $\mathbb E [X\vert S] = g(S)$. That is, $E[X\vert S]$ is the function $g$ evaluated at the realisation of the random variable $S$.
Notice that this is different from the expectation of $X$ conditional on the event $\{S=S\}$, which would just be equal to the unconditional expectation of $X$.
If you are familiar with more general definitions of conditional expectation, you can verify that the definition I’ve given above is equivalent in the setting you’ve described.