 Solution

Measure each generator $S_j$. If a Pauli error $E$ has occurred, its outcome is $(-1{)}^{s_j}$, whereThe bit vector $s=(s_1,\dots ,s_m)$ is the error syndrome. Choose one representative $E_s$ with this syndrome. Another Pauli $E$ has the same syndrome exactly when $E_s^{†}E$ commutes with every $S_j$, namely when $E_s^{†}E\in C(S)$. ThereforeThis stabilizer-syndrome coset is all the syndrome reveals: multiplication by a stabilizer changes nothing on the code, while multiplication by an element of $C(S)\setminus S$ can change the logical state without changing any syndrome bit.