Disclaimer: Google search produces a lot of references closely related to my question so I think it should be easy for me to find out the answer on my own, unfortunately due to a complete lack of familiarity with the subject, and due to not being a math student in general, I am having a hard time spotting the answer.
Question: I have found the following two facts in literature:
1) Given a compact connected Lie group $G$, the Lie algebra cohomology of $\mathrm{Lie}(G)$ with values in the module $\mathbb{R}$ with trivial action is isomorphic to the de Rham cohomology of the real valued left invariant forms on $G$: $$ H^\bullet(\mathrm{Lie}(G);\mathbb{R}) \cong H^\bullet_L(G;\mathbb{R})\,.$$ 2) The de Rham cohomology of the left invariant forms is isomorphic to the ordinary de Rham cohomology: $$H^\bullet_L(G;\mathbb{R}) \cong H^\bullet(G;\mathbb{R})\,.$$
My question is, is there a "left invariant" de Rham cohomology isomorphic to the Lie algebra cohomology valued in some arbitrary module? If yes, is that "left invariant" de Rham cohomology isomorphic to some ordinary de Rham cohomology?
Part (2) is more complicated and geometric, but as for part (1), the complex of invariant forms just coincides with certain standard complex which is used to calculate $H^\bullet (\mathfrak{g}, M)$, the Chevalley-Eilenberg complex. It is constructed for any $\mathfrak{g}$-module $M$. See section 7.7 in Weibel's "Introduction to homological algebra".