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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.

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This is not an answer per se, 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): 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.