The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter case, we also get an injection $S\to\beta$, by taking any element $S$ to the least element in its preimage. In either case, we see $S$ is well-orderable. Thus both (I) and (S) imply well-ordering theorem and hence AC.

The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter case, we also get an injection $S\to\beta$, by taking any element $S$ to the least element in its preimage. In either case, we see $S$ is well-orderable. Thus both (I) and (S) imply well-ordering theorem and hence AC.

Since both (I) and (S) are equivalent to AC, they are also equivalent to each other.