3 Added distinction between finite/infinite sets

It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary operation of symmetric difference forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice.

So my question is

Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation $\ast$ on $S$ making $(S,\ast)$ into a group?

2 Changed * to \ast, as markdown doesn't seem to like it

It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: the set of finite subsets of $S$ with the binary operation of symmetric difference forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice.

So my question is

Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation  \ast$on$S$making$(S,)$(S,\ast)$ into a group?

1