As stated above, $S$ is polynomial ring, and since the polynomial ring is $S$ and $S(-d)$ are finite over $S$ as graded modules, we can say that $\operatorname{Hom}(S(-d),S)$ is also graded.
My question is how can we show that $\operatorname{Hom}(S(-d),S)\cong S(d)$?
best regards
$S$ can be any commutative graded ring. Recall that for two graded $S$-modules $M,N$ the graded hom (or internal hom) $\underline{\hom}(M,N)$ is given by $$\underline{\hom}(M,N)_n := \hom(M,N[n]) \subseteq \prod_p \hom(M_p,N_{p+n}).$$If $M$ is of finite presentation, then we have $\bigoplus_n \underline{\hom}(M,N)_n = \hom(U(M),U(N))$, where $U : \mathsf{grMod}(S) \to \mathsf{Mod}(S)$ is the forgetful functor (which is forgotten by most people), but we don't need this assumption to talk about the graded module $\underline{\hom}(M,N)$. (The other answers focus on $\hom(U(M),U(N))$, written incorrectly as $\hom(M,N)$, but it was explicitly asked for the graded hom.) I suggest to think $U$ not as a forgetful functor at all, and not even think of graded $S$-modules as special $S$-modules. "Graded" is not a property, it is an extra structure, but not on $S$-modules, but rather on graded abelian groups. Also notice that when talking about (what is usually called) $S$-modules we actually mean $T$-modules, where $T$ is the "underlying" commutative ring of $S$. Actually, since $S$ is just a commutative algebra object in the tensor category of graded abelian groups, $S$-modules coincide by definition with which are usually called graded $S$-modules! (See also Remark 5.4.12 in my thesis.)
After this long digression, here is the answer to the question: Observe that for every graded $S$-module $M$ we have $\underline{\hom}(S,M) \cong M$. In degree $p$, this isomorphism is given by $\hom(S,M[p]) \cong M_p$, which evaluates at $1 \in S_0$. Then, we have $$\underline{\hom}(S[-d],M) = \underline{\hom}(S,M[d]) \cong M[d].$$