Let me make sure I understand the question: $V$ is spanned by all functions of the form $g \mapsto f(gh)$, $W$ is spanned by the functions $g \mapsto v_i(g)$, it is clear from the given formula that $W \subseteq V$ and you are asking whether $V=W$. Also, I am assuming you are dealing with reduced algebraic groups over a field $k$, since otherwise what you mean by an orbit is a bit confusing. So I can think of elements of $\mathcal{O}(G)$ as functions on the $\overline{k}$ points of $G$.
Yes. Since the $e_i$ are a basis for $\mathcal{O}_G$, in particular $e_1$, $e_2$, ..., $e_m$ are linearly independent. So we can find $h_1$, $h_2$, ..., $h_m$ in $G(\overline{k})$ so that the matrix $\left( e_i(h_j) \right)$ is invertible.
Inverting the equations $$f(g h_j) = \sum v_i(g) e_i(h_j)$$ we see that $g \mapsto v_i(g)$ is in the span of functions of the form $g \mapsto f(gh)$, as desired.