I know, basically, two answers to this question.  The first one starts from the observation that what you call "a relatively easy theorem" generalizes to infinite-dimensional Lie algebras $L$ by replacing the exterior algebra generated by $L^\ast$ with the exterior coalgebra cogenerated by $L$.  Then the construction becomes covariant, but produces a (conilpotent) DG-coalgebra rather than a DG-algebra.

Now it is just a fact that the natural equivalence relation on DG-coalgebras is more delicate than the quasi-isomorphism.  There is a certain class of "filtered quasi-isomorphisms" that one is supposed to invert when dealing with (conilpotent) DG-coalgebras.  The map of Chevalley-Eilenberg complexes that you describe, when considered as a map of the homological, rather than cohomological, Chevalley-Eilenberg complexes, is not a filtered quasi-isomorphism of DG-coalgebras and cannot be obtained from such using compositions and fractions.  This saves the derived Koszul duality as an equivalence between appropriate localizations of the categories of (augmented) DG-algebras and (conilpotent) DG-coalgebras.

The second answer describes what happens if one insists on considering the cohomological Chevalley-Eilenberg complex of a finite-dimensional Lie algebra, viewed as a DG-algebra up to a quasi-isomorphism.  One can also apply the derived Koszul duality to it, obtaining a certain DG-coalgebra.  This DG-coalgebra is, basically, the k-linear dual to the complete topological DG-algebra of derived adic completion of the enveloping algebra U(L) with respect to its augmentation ideal.  So if a morphism of finite-dimensional Lie algebras induces an isomorphism of cohomology, this should be understood to mean that it induces an isomorphism of the derived adic completions of the enveloping algebras at their augmentation ideals.  Admittedly, this is not a geometric interpretation.