Hint: Use the fact that, for every integer valued random variable $S$ and every integer $x$,
$$
\mathrm P(S=x)=\int_0^{1}\mathrm E(\mathrm e^{2\pi\mathrm itS})\cdot\mathrm e^{-2\pi\mathrm itx}\,\mathrm dt,
$$
and the fact that, in the present case,
$$
\mathrm E(\mathrm e^{2\pi\mathrm itS_n})=\prod_{k=1}^n(1-p_k+p_k\mathrm e^{2\pi\mathrm ita_k/n}).
$$
Edit: The second identity above is a consequence of the definition of $\mathrm E(\mathrm e^{2\pi\mathrm itS_n})$ and of the joint distribution of the random variables $(X_k)_{1\leqslant k\leqslant n}$.
The first identity is an application of the general principle that integrating a discrete sum of complex exponentials against the conjugate of a complex exponential extracts the coefficient of the corresponding exponential from the sum. Namely, for every integers $x$ and $y$,
$$
\int_0^{1}\mathrm e^{2\pi\mathrm ity}\cdot\mathrm e^{-2\pi\mathrm itx}\,\mathrm dt=[x=y],
$$
hence, for every distinct integers $x_k$ and every coefficients $p_k$,
$$
p_\ell=\int_0^{1}\left(\sum_kp_k\mathrm e^{2\pi\mathrm itx_k}\right)\cdot\mathrm e^{-2\pi\mathrm itx_\ell}\,\mathrm dt.
$$
Applying this to the integer valued random variable $S$ such that $p_k=\mathrm P(S=x_k)$ yields the first formula above.
When $S$ is not integer valued, use the fact that, for every real numbers $x$ and $y$,
$$
\lim_{N\to\infty}\int_0^1\mathrm e^{N\mathrm ity}\cdot\mathrm e^{-N\mathrm itx}\,\mathrm dt=[x=y],
$$
hence, for every discrete random variable $S$,
$$
\mathrm P(S=x)=\lim_{N\to\infty}\int_0^1\mathrm E(\mathrm e^{N\mathrm itS})\cdot\mathrm e^{-N\mathrm itx}\,\mathrm dt.
$$