Let $p$ be a prime divisor of $n$ and pick $S$ to be the set of all numbers with all prime factors $> p$. This works, since no prime at most $p$ can divide $\frac{a^c - b^c}{a-b}$. Suppose not, and say $q$ divided it, then $a^c \equiv b^c \pmod q$, write $a \equiv g^{m} \pmod q$ and $b = g^{n} \pmod q$ for some primitive root $g$, then we have $cm \equiv cn \pmod {q-1}$, but since $c$ is coprime to $p-1$, we must have $m \equiv n \pmod {p-1}$ and so in fact, $q | a-b$, but then $v_q(a^c - b^c) = v_q(a-b) + v_q(c) = v_q(a-b)$ so $\frac{a^c - b^c}{a-b}$ is not divisible by $q$. Finally, since $n$ does have a prime factor at most $p$, it is not in $S$, so we are done. $\blacksquare$