If sat(H) contains x, then x/y = a/b (in H/S), and ubx = uay. Thus something in S carries x into H. Conversely, if ux = a, for some a in H, then xw/w = a/u, and x is in sat(H). Since HS lies in H, sat(H) contains H.
If H and J are ideals, use the above criteria to show sat(H∩J) = sat(H)∩sat(J). One direction is obvious, so assume ux lies in H and vx lies in J. Now uvx lies in H and J, and we're done.
If H contains J, sat(H) contains sat(J).
Sat(H) is the entire ring R, iff 1 times something in S lies in H, iff H and S intersect.
Take the saturation of H with respect to S, then saturate this ideal with respect to another closed set T. Now x is in sat(sat(H)) when vux lies in H, for some v in T and some u in S. Cross multiply all the elements of S by all the elements of T to get a multiplicatively closed set Z, which includes uv. Thus x is in the saturation of H with respect to Z. By symmetry, saturation with respect to S, then T, is the same as saturation with respect to T, then S. Both are equal to saturation with respect to Z.