The Symmetric group is already known to be a group so you do not need to show associativity. In general to show a subset H is a subgroup you need:
1) The inverse is in H
2) H is closed under the group operation (a,b in h implies ab in H).
3) If a is in H then a-1 is in H
however 2 and 3 imply 1 so you only really need to show the last two. The one step solution is to show that a in H, b in H implies ab-1 is in H because that implies 2 and 3. In your case let f,g be bijections from S to S. Then show:
1) f(a) = a implies f-1(a) =a
2) f(a) = a, g(a) =a implies f(g(a)) = a