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): 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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