I believe you are missing an $n$ in your definition of $K_n(t)$, that is $K_n(t)=\log E(e^{tnS_n})$. I assume in the sequel that this is what you meant.

If $S_n$ satisfies the large deviations principle with a non-convex rate function, then clearly the rate function is not $K^*$. So your question  can be rephrased as "does there exist $S_n$ that satisfies both your assumptions and the LDP with non-convex rate function?"

Here is an example: take $S_n=0$ with probability $1/2$ and $S_n=1$ with probability $1/2$. Then $K_n(t)=[\log ( (e^{tn}+1)/2)]$.
Thus, $K(t)=t$ for $t\geq 0$ and $K(t)=0$ for $t<0$. You get $K^*(a)=\infty$ for $a<0$, $K^*(a)=0$ for $a\in [0,1]$ and $K^*(a)=\infty$ for $a>1$. But
$P(S_n\in (0,1))=0$ which contradicts a LDP with rate $K^*$ (the lower bound fails). In fact, you get a LDP with rate function $I$ satisfying  $I(a)=\infty$ if $a\notin \{0,1\}$ and $I(0)=I(1)=0$.

Edit: with the original statement of the OP, it is $S_n/n$ that satisfies the LDP, not $S_n$, and then the same example works with the case $S_n=1$ replaced by $S_n=n$.