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.
EDIT: I have been asked in a comment what a filtered quasi-isomorphism of DG-coalgebras is, so let me add an explanation of this to the answer. This notion does not presume a fixed filtration on a DG-coalgebra, but rather a class of admissible filtrations. We consider coaugmented DG-coalgebras, i.e., the basic field $k$ is embedded as a subcoalgebra into a DG-coalgebra $C$ and its image is annihilated by the differential. An increasing filtration $F$ on $C$ is admissible if $F_0C=0$, $F_1C=k$, the components $F_iC$ are preserved by the differential, and the filtration $F$ is compatible with the comultiplication.
Not every coalgebra admits an admissible filtration at all! A coalgebra is called conilpotent if it has an admissible filtration (just as a coalgebra, let us drop the condition of compatibility with the differential for a moment). In this case, it has a certain maximal admissible filtration (which is preserved by the differential automatically). However, filtered quasi-isomorphisms of conilpotent DG-coalgebras are defined in terms of arbitrary admissible filtrations. A morphism of DG-coalgebras is a filtered quasi-isomorphism if admissible filtrations can be chosen on its source and target so that the morphism is compatible with them and is a quasi-isomorphism on all filtration components.
In Donald Kahn's 1966 paper "On stable homotopy modules" (Inventiones 1, pp 375–379, doi:10.1007/BF01389739)
he constructs examples of CW complexes with 3 cells, whose stable homotopy groups are not finitely generated as modules over the ring of stable homotopy groups of spheres. I believe this answers in the negative a stable analogue of the question.
Perhaps a Freudenthal suspension theorem argument, together with the stable result, can imply the desired unstable result. I am sorry but I do not have a few minutes to spare to explore this possibility; perhaps others are interested enough to try it out, though. (I am sorry if this incomplete answer is inappropriate for this site, which I am a beginner to. Please feel free to delete this answer if so. I would certainly have made this all a comment rather than an answer, but don't have enough points on this site to leave comments.)
Best Answer
Another counterexample: let $A$ be the algebra $\mathbb{Q}[y,z]/(y^2) \otimes \bigwedge(x)$ with $x$ in degree 1, $y$ and $z$ in degree 2. Put a differential on this by $z \mapsto xy$. This is a commutative dga in characteristic 0 generated in positive degrees, but of course it's not free. Its homology is spanned by the classes $xz^i$ and $yz^i$ for all $i \geq 0$, and the product on the homology algebra is trivial. So it is infinitely generated as an algebra.