Proof of that in an integral domain, every prime element is irreducible. I would like to prove that in an integral domain $R$, every prime element $p$ is irreducible. I understand the case where $p = ab$ but the textbooks I have read do not address the case where $p \neq ab$, i.e.,$px = ab$.

I was wondering why they do not discuss the case where $px = ab$.
You don't need to examine the case $p\neq ab$.

In order to prove every prime element $p$ is irreducible you have to show IF $p=ab$, then $a$ or $b$ is an unit (see the definition of irreducible element). 

However, if we have $p\neq ab$? It doesn't matter, we don't care, what matters to us is just the case whenever $p=ab$.
Proof:
suppose P is a prime element of an integral domain R, and p=ab
we know that, P|P => P|ab => P|a or P|b
              P|a and p=ab or a|P => a is unit 
similarly P|b and p=ab ir b|p => b is unit,
so either a is a unit or b is a unit, hence p must be irreducible element,