I don't know how to show the claim from duality theory but it follows quite easily from the formula $\omega_B=Hom_A(B,\omega_A)$. Here, $A$ is a CM-ring and $B$ is an $A$-algebra which is a free $A$-module of finite rank.

In our case, we take $B=k[S]$ and $A=k[t^f]$ where $f\in\mathbb N$ is larger than the maximal gap of $S$. Then $f,f+1,f+2,\ldots\in S$. Let $I:=\{0,1,\ldots,f-1\}$. Then $B$ is a free $A$-module with basis $t^M:=\{t^n\mid n\in M\}$ where $M=(I\cap S)\cup(f+(I\setminus S))$. Since $\omega_A$ is freely generated by an element of degree $f$ (namely $d\ t^f)$) the canonical module of $B$ is a free $A$-module with basis $t^{M'}$ where
$$
M'=f-M=(f-(I\cap S))\cup(-(I\setminus S))
$$
From
$$
-M'=(-f+(I\cap S))\cup(I\setminus S)
$$
one easily checks that $t^{-M'}$ is a basis of $k[\mathbb Z\setminus S]$ as a $k[t^{-f}]$-module.

This identifies $\omega_B$ with $k[t^{-n}\mid n\in\mathbb Z\setminus S]$, at least as an $A$-module. As a last step one has to verify that this is an isomorphism of $B$-modules which is elementary.