Prime ideals of a finite direct product ring

If I am not mistaken, you've proven ideals of that form are prime, and now you are headed for the converse. Let $I$ be a prime ideal of $R$.

I assume you already know the ideal structure of finite products of rings, so you can see why the ideal must be of the form $I=\prod I_i$ for some ideals $I_i\lhd R_i$. Then $R/I=(\prod R_i)/(\prod I_i)\cong\prod R_i/I_i$.

Now, in order for this product to be a prime ring (which is exactly what a quotient by a prime ideal gives you) it is necessary for all factors to be zero except one, and the one that is nonzero must also be a prime ring.

The first condition implies there is a $j$ such that $R_k=I_k$ for all $k\neq j$, and the second condtion implies $I_j$ is a prime ideal of $R_j$.

A quick note about KonstantinArdakov's comment-solution: it's fine, except for a small danger of readers misunderstanding part of the logic.

As we know, the general definition of prime ideals says that $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$ for any two ideals $A,B$ of $R$. That is, it's the ideal-wise version of the element-wise condition for primality. So something needs to be said about why $e_ie_j=0$ implies one of the two is in $P$.

Now, given that at least one of $a$ and $b$ is central, we can conclude that $(a)(b)=\{0\}\subseteq P$, and it follows that $a\in P$ or $b\in P$, so everything works as hoped.