You are probably stuck because the random variable $N_n$ may assume far too many values for Kolmogorov's inequality to provide an effective upper bound. This suggests to deal separately with the case when $N_n$ is around $a_n$ (which, by Kolmogorov's inequality, should yield small values of $S_{N_n}-S_{a_n}$) and with the case when $N_n$ is far from $a_n$ (which, from the hypothesis that $N_n/a_n\to1$ in probability, should have small probability).
Hence, let us introduce, for a given positive $\varepsilon$, the event $A_n=[(1-\varepsilon) a_n\leqslant N_n\leqslant (1+\varepsilon) a_n].$ On the one hand, $N_n/a_n\to1$ in probability hence $A_n$ is typical in the sense that $\mathrm P(\Omega\setminus A_n)\to0$.
On the other hand, $|S_{N_n}-S_{a_n}|\leqslant |S_{N_n}-S_{(1-\varepsilon) a_n}|+|S_{a_n}-S_{(1-\varepsilon) a_n}|$ hence, on the event $A_n$, $ |S_{N_n}-S_{a_n}|\leqslant 2M_n,\qquad M_n=\sup\limits_{1\leqslant k\leqslant 2\varepsilon a_n}|T_k|,\qquad T_k=S_{(1-\varepsilon) a_n+k}-S_{(1-\varepsilon) a_n}. $ Now, we are back to the realm where Kolmogorov's inequality applies, and yields $ \mathrm P(M_n\geqslant x\sqrt{a_n})\leqslant (a_nx^2)^{-1}\mathrm{Var}(T_{2\varepsilon a_n})=(a_nx^2)^{-1}(2\varepsilon a_n)\sigma^2=2\varepsilon x^{-2}\sigma^2. $ Putting our two estimates together yields $ \mathrm P(|S_{N_n}-S_{a_n}|\geqslant 2x\sqrt{a_n})\leqslant\mathrm P(\Omega\setminus A_n)+\mathrm P(M_n\geqslant x\sqrt{a_n})\leqslant\mathrm P(\Omega\setminus A_n)+2\varepsilon x^{-2}\sigma^2. $ This proves that, for every positive $\varepsilon$, $ \limsup\limits_{n\to\infty}\ \mathrm P(|S_{N_n}-S_{a_n}|\geqslant 2x\sqrt{a_n})\leqslant2\varepsilon x^{-2}\sigma^2, $ hence $\mathrm P(|S_{N_n}-S_{a_n}|\geqslant2x\sqrt{a_n})\to0$ for every $x$, that is, $S_{N_n}/\sqrt{a_n}-S_{a_n}/\sqrt{a_n}\to0$ in probability.
By the usual central limit theorem, since $a_n\to+\infty$, $S_{a_n}/\sqrt{a_n}$ converges in distribution to a centered gaussian distribution with variance $\sigma^2$, hence $S_{N_n}/\sqrt{a_n}$ converges in distribution to the same centered gaussian distribution with variance $\sigma^2$.