I guess you are asking for feedback on your proof.
For the implication in this direction ==>, you write that 1 is an additive generator for A. But there may be elements in A that cannot be written as a finite sum of 1's. Instead, remember that an ideal is closed under multiplication with elements from the ring. So since 1 is in the ideal, then for all r in A, we have also that r = r1 is in the ideal (so all elements of A are in I, hence I=A).
The other direction is good.
As you yourself note, this really hasn't got much to do with prime ideals. Rather, it is true that the only ideal containing 1 is the entire ring - no proper ideals contain the identity.