Let f: R -> R' be an isomorphism. If R is commutative, and a', b' are elements of R', then a' = f(a) and b' = f(b) for some a,b in R. Then we have: a'b' = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = b'a', so R' is commutative. Conversely if R' is commutative, and a,b are elements of R, then f(a) = a' and f(b) = b' for some a',b' in R'. Then we have: f(ab) = f(a)f(b) = a'b' = b'a' = f(b)f(a) = f(ba), so f(ab) = f(ba), and by injectivity of f, ab = a. Hence R is commutative.
A quick note: isomorphism is the "right" notion so that any nontrivial algebraic property carries over isomorphism. If you can say something in algebraic language, like "R is commutative," it's a safe bet that the same holds for isomorphic rings. You don't have to go proving these things all the time.