Is the separability of the space needed in the proof of the Prohorov's theorem?
Separability is not necessary. In fact, tightness of a family of Borel probability measures implies relative compactness in the vague/weak-* topology on any completely regular space. For instance, this can be found in volume 4 of Fremlin's Measure Theory. Specifically Proposition 437U (b) shows that tight families are compact in the narrow topology, and 437K (c) shows that for completely regular spaces, the narrow topology agrees with the weak-* topology.
My original answer below answers the wrong question - the question is about whether tight implies relatively compact, rather than the other way.
Let $\kappa$ be a real-valued measurable cardinal, and $\mu : \mathcal{P}(\kappa) \rightarrow [0,1]$ a probability measure vanishing on singletons. Consider $\kappa$ to be a discrete metric space. Then the 1-element family $\{\mu\}$ is compact, because it is a singleton, but it is not tight because all compact subsets of $\kappa$ are finite sets, so have measure zero.
You say that completeness is not necessary, but (unless you are making an extra assumption) it is, for essentially the same reason - there are separable metric spaces with Borel measures on them that are not tight.
You are correct that separability is not needed. However, there is also not really any loss of generality in assuming it. For suppose that $\Pi$ is tight. Then for every $n$ there exists a compact set $K_n$ such that $\mu(K_n) > 1-\frac{1}{n}$ for all $\mu \in \Pi$. So if we set $S_0 = \bigcup_{n=1}^\infty K_n$, then $S_0$ is separable and $\mu(S_0) = 1$ for all $\mu \in \Pi$. We can now view $\Pi$ as a set of probability measures on $S_0$, and it is still tight (since the $K_n$ are also compact in $S_0$). The separable case of the theorem then implies that $\Pi$ is weakly relatively compact in $\mathcal{P}(S_0)$, i.e. every sequence in $\Pi$ has a subsequence converging weakly in $\mathcal{P}(S_0)$, and you can easily check that such a subsequence also converges weakly in $\mathcal{P}(S)$. So $\Pi$ is weakly relatively compact in $\mathcal{P}(S)$, as desired.
In other words, once you have a tight family, then all those measures live on a separable subset of $S$ anyway, so the rest of the space is irrelevant and might as well not be there.