The answer is $a = 3, b = 1, c = 2$ and $a = b \in \mathbb{Q}^+, c = 1$.
Applying induction on $n$ by using Bernoulli's inequality gives
$$\frac{n^{r+1}}{r+1} \le S_r(n) \le \frac{(n+1)^{r+1}}{r+1}$$
for all positive integer $n$ and positive rational number $r$.
As $S_a(n) = (S_b(n))^c$ letting $r = a$ and $r = b$ gives that
$$\frac{n^{a+1}}{a+1} \le \left( \frac{(n+1)^{b+1}}{b+1} \right)^c \text{ and } \left( \frac{n^{b+1}}{b+1} \right)^c \le \frac{(n+1)^{a+1}}{a+1}$$
i.e.
$$\frac{n^{(b+1)c}}{(n+1)^{a+1}} \le \frac{(b+1)^c}{a+1} \le \frac{(n+1)^{(b+1)c}}{n^{a+1}}$$
holds for infinitely many positive integers $n$. By letting $n \to \infty$ in the last inequality, we obtain that $(b+1)c = a+1$ and $(b+1)^c = a+1$. If $c=1$, then $a=b$ and we get the trivial solutions.
If $c > 1$, then $c = (b+1)^{c-1}$ implies that $b$ is an integer since $c$ is an integer. As $b \ge 1$, we get that $c \ge 2^{c-1}$ and hence $c = 2$. This leads to $b = 1, a = 3$ and this solution clearly satisfies the condition.