Well, there exists a homomorphism $\phi: R \mapsto R'$. It just so happens that this is an isomorphism, but that is not reeeeeeeeally needed. You need surjectivity, but not injectivity.
So, as this is an isomorphism we have that every element of $R'$ is of the form $a\phi$ for some $a \in R$. Now, use the fact that this is a homomorphism, so $(a\phi)(b\phi) = \ldots$.