Order the ideals in R-S that contain H by inclusion, and use zorn's lemma to find a maximal ideal P.
Suppose P is not prime, with A*B as counterexample. The ideals P+A and P+B properly contain P, and must intersect S. Select u and v from P, and x from A, and y from B, such that u+x and v+y lie in S. Since S is closed under multiplication, the product of these two elements lies in S. Since P contains xy, the product also lies in P, which contradicts P disjoint from S.
Note, if S were empty, P could be all of R, which is technically not a prime ideal. We can always set S to 1, hence maximal ideals are prime. Start the chain at 0, and there is at least one maximal, prime ideal.
If R does not contain 1, we don't have the luxury of setting S to 1, hence there may be no prime ideals. Let R contain x, such that x2 = 0. Now R consists of 0, x, 2x, 3x, etc, and their opposites, and that's all. If the ideal P is prime it contains 0, and x, and all of R. There are no prime ideals.
The ideal generated by 2x is maximal, but not prime.
A ring R contains at least one prime ideal iff there is an x such that the powers of x are never 0.