In "Weierstrass-Stone, the Theorem" by Joao Prolla, there is a Stone-Weierstrass theorem for modules, stated as the following:

Let $\mathcal{A}$ be a subalegebra of $C(X, \mathbb{R})$ and $(E, \|\cdot\|)$ be a normed space over $\mathbb{R}$. Let $W\subset C(X, E)$ be a vector subspace which is an $\mathcal{A}$-module. For each $f\in C(X, E)$ and $\epsilon>0$, there exists $g\in W$ such that $\|f-g\|<\epsilon$ if and only if for each $x\in X$, there exists $g_x\in W$ such that $\|f(t) - g_x(t)\| < \epsilon$ for all $t\in [x]_{\mathcal{A}}$, where $[x]_\mathcal{A}$ is the equivalent class of $x$ under $\mathcal{A}$.

I know that the above theorem can be extended to $\mathcal{A}\subset C(X, \mathbb{C})$ with $\mathcal{A}$ being a self-adjoint subalgebra. I wonder whether there are some similar results for modules of non-self-adjoint algebras.

I'm interested in generalizing the above theorem into the following case. Let $\mathcal{S}$ be a finite subset of $C([0, 1], E)$, denoted as $S:=\{s_1, \ldots, s_m\}$, and $\mathcal{A}\subset C([0, 1], \mathbb{C})$ be a subalgebra (not necessarily self-adjoint). Then $W := \mathrm{span}\{as : a\in \mathcal{A}, s\in \mathcal{S}\}$ is a vector subspace which is an $\mathcal{A}$-module. Shall we still claim that $f\in \overline{W}$ if and only if $f\big\vert_{[x]_{\mathcal{A}}} \in \overline{W}\big\vert_{[x]_{\mathcal{A}}}$? Is there any counter-example to this statement? Or is it an open problem in general?

Note: For any $x\in X$, the equivalent class $[x]_{\mathcal{A}}$ is a subset of $X$ such that $\forall u, v\in [x]_{\mathcal{A}}$, we have $a(u) = a(v)$ for all $a\in \mathcal{A}$.