This is not an answer per se [Edit: OK, maybe it is! I was a little fuzzy on exactly what was being asked for when I wrote this, and in the past Martin has expressed unhappiness with responses which he feels have not answered his questions.] but it should be useful for those who are thinking about the problem (c.f. Kevin Buzzard's answer) to know the following classic result.
Theorem (Schreier-Ulam, 1933): Schreier-Ulam): The only nontrivial proper normal subgroups of $S_{\infty}$ are $\mathfrak{s}_{\infty} = \bigcup_{n \geq 1} S_n$ and $\mathfrak{a}_{\infty} = \bigcup_{n \geq 1} A_n$, i.e. the "little symmetric group" of all permutations which move only finitely many elements and its index two alternating subgroup.
Reference: J. Schreier and S. Ulam, Über die Permutationsgruppe der natürlichen Zahlenfolge. Stud. Math. 4, 134-141 (1933).
Addendum: Certainly this theorem implies that any homomorphism from $S_{\infty}$ into a group $G$ which restricts to the sign homomorphism on $\mathfrak{s}_{\infty}$ must have kernel precisely equal to $\mathfrak{a}_{\infty}$. Whether this answers the question depends, I suppose, on how much you care about what the induced monomorphism $S_{\infty}/\mathfrak{a}_{\infty} \hookrightarrow G$ looks like.
