Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space and $X:\Omega\rightarrow \mathbb{R}$ a random variable on it.
Now the definition of expected value would be $$\mathbb{E}[X]=\int _{\Omega} Xd\mathbb{P}.\quad(1)$$ This is true in total generality.
But if $\mathbb{P}$ is the counting measure and $\Omega$ is countable and $\mathcal{A}$ contains all subset of $\Omega$, then $(1)$ reduces to $$\mathbb{E}[X]= \sum_{\omega \in {\Omega}} \mathbb{P}(\omega)X(\omega),\quad(2)$$ which is a formula we know from elementary probability theory.
My question is: Does $(2)$ also hold in certain cases for other probability spaces; is there perhaps a general criterion about $(\Omega,\mathcal{A},\mathbb{P})$ and $X$ that tells me when I can use $(2)$ ?
Firstly, a slight correction. The counting measure is not a probability measure unless $\Omega$ is a singleton. You probably just mean that $\mathbb{P}$ is any probability measure on a countable space. $\mathbb{P}$ is automatically absolutely continuous with respect to the counting measure and so has a density against the counting measure. It's then easy to see that density just has value $\mathbb{P}(\omega)$ at $\omega$ and hence $\mathbb{E}[X] = \sum_{\omega \in \Omega} X(\omega) \mathbb{P}(\omega)$.
Of course, starting with such a measure $\mathbb{P}$ on a countable $\Omega$, one can do silly things by adding to $\Omega$ an uncountable set that you declare to have $0$-measure. That is, define $\Omega', \mathbb{P'}$ by $\Omega' = \Omega \cup \tilde{\Omega}$ for some uncountable $\Omega$ and defining $\mathbb{P'}$ by $\mathbb{P'}(A) = \mathbb{P}(A)$ for $A \subseteq \Omega$ and $\mathbb{P'}(\Omega') = 0$. Then $(\Omega', \mathbb{P'})$ has the desired property.
In fact, this is really all you can do. That is, having the discrete definition of expectations is equivalent to being in the elementary case, up to a set of measure $0$.
Proof: We have for measurable $A \subseteq \Omega$, $\mathbb{P}(A) = \int_\Omega 1_A d \mathbb{P} = \sum_{\omega \in A} \mathbb{P}(\omega)$. Since $\mathbb{P}$ is a finite measure, this means that there can only be countably many $\omega \in \Omega$ such that $\mathbb{P}(\omega) > 0$ (since uncountable sums of strictly positive terms diverge). Define $\Omega_1 = \{ \omega \in \Omega: \mathbb{P}(\omega) > 0\}$ and $\Omega_2 = \Omega \setminus \Omega_1$. Then by considering $X = 1_{\Omega_2}$, $$\mathbb{P}(\Omega_2) = \sum_{\omega \in \Omega_2} \mathbb{P}(\omega) = 0$$ by definition of $\Omega_1$.
Note: Throughout I have ignored the choice of $\sigma$-algebra. Obviously for $\mathbb{P}(\omega)$ to make sense for every $\omega$, you need $\mathcal{A}$ to be the discrete $\sigma$-algebra so I restrict to this case.