Let $R = \mathbb{C}[x,y]$ (or possibly $R = \mathbb{C}[[x,y]]$, although I think we don't need the local hypothesis, and can work in a graded setting instead) and let $G \leqslant \operatorname{GL}(2,\mathbb{C})$ be a finite reflection-free subgroup. Let $S = R^G$ be the invariant ring, which is Cohen-Macaulay. It is well-known that the indecomposable maximal Cohen-Macaulay (MCM) modules over $S$ are in bijection with the irreps of $G$; more precisely, if $G$ has pairwise non-isomorphic irreps $V_0, \dots, V_n$ (where we'll assume that $V_0$ is the trivial representation) then the indecomposable MCM modules are given by \begin{align*} M_i := (R \otimes V_i)^G. \end{align*} In particular, $M_0 = S$.
Now consider the MCM module $K_S = (R \otimes V_{\det})^G$, where $V_\det$ is the determinant representation. This is the so-called canonical module of $S$, which implies in particular that $\operatorname{Ext}^i_S(M,K_S) = 0$ for all $i \geqslant 1$ and for all MCM $S$-modules $M$. My question is then:
Let $K_S = (R \otimes V_{\det})^G$. How can one show that $\operatorname{Ext}^i_S(M,K_S) = 0$ for all $i \geqslant 1$ and for all MCM $S$-modules $M$ "directly", i.e. without reference to the fact that $K_S$ is the canonical module?
Edited to add: I know that the module $K_S$ fits into two exact sequences, namely \begin{align*} &0 \to K_S \to E \to \mathfrak{m} \to 0, \\ &0 \to K_S \to E \to S \to \mathbb{C}_{\geqslant 1} \to 0, \end{align*} (obviously these are essentially the same exact sequence). I've been trying to exploit this fact to prove the result.
Incidentally, I don't have a good modern-day reference for the fact that the module $(R \otimes V_{\det})^G$ is the canonical module of $S$, so a reference would be appreciated.