$\mathfrak g$ be a Lie algebra (if it matters, right now I only care about finite-dimensional Lie algebras in characteristic $0$, although I'm never opposed to hearing about more general cases). Recall that it determines a differential graded algebra ("the complex that computes Lie algebra cohomology"), with $k$th component
$\wedge^k \mathfrak g^*$ and differential determined by the bracket. It is a complex by the Jacobi identity. Thinking in terms of supergeometry, I will call this dga $\mathcal C^\infty([-1]\mathfrak g)$.
Moreover, if $\mathfrak h \to \mathfrak g$ is a homomorphism of Lie algebras, then we get a homomorphism of dgas $\mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$, so that $\mathcal C^\infty\circ [-1]$ is a contravariant functor. The dga is a complete invariant: the functor is full and faithful.
Suppose that $\mathfrak h,\mathfrak g$ are two Lie algebras and $f: \mathfrak h \to \mathfrak g$ a Lie algebra homomorphism. Suppose furthermore that the corresponding map $f^*: \mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$ is a quasi-isomorphism, i.e. it induces an isomorphism on cohomology. Does it follow that $f$ is an isomorphism?
When I ask it this way, it sounds strongly like the answer should be "no": almost never is cohomology a complete invariant. For example, the cohomology in degree $1$ sees only the abelianizations of $\mathfrak g,\mathfrak h$. But on the other hand, research I'm doing on Lie algebroids suggests that the similar statement with "algebra" replaced by "algebroid" throughout should be true. I don't see a direct proof even in the "algebra" case, but I feel like there should be either a trivial counterexample or an easy argument in favor. In either case, though, and maybe because it's late at night, I'm stuck.
Which is all to say that secretly I care about algebroids, so if any of y'all know a good reference for the problem at that generality, please send it my way. But I will happily accept an answer just for algebras if one is provided.