some things to clean up: first, assume R is NOT the 0-ring (because that IS a domain, because there are no zero-divisors, indeed no non-zero elements).

then we have a ≠ 0 in R. then we can define f:R to R by f(r) = ar, and f is not the 0-map, since aa ≠ 0 (because R is a domain).

as you have it, we then know that if f(r) = f(s), then ar = as, so ar - as = 0, so a(r - s) = 0, thus r - s = 0, so r = s.

since R is assumed finite, f injective = f bijective. so there is some w in R with f(w) = a.

now, prove that w must be the identity of R (you'll need to use commutativity of R).