If $\mathfrak{g}$ is a Lie $R$-algebra, then the Chevalley-Eilenberg complex defines the cohomology modules $H^k(\mathfrak{g})$. If $H^\ast(\mathfrak{g})=\bigoplus_kH^k(\mathfrak{g})$, then the diagonal map $\mu\!:\mathfrak{g}\longrightarrow\mathfrak{g}\!\times\!\mathfrak{g}$ and zero map $\iota\!:\mathfrak{g}\!\longrightarrow\!0$ induce maps $$\mu^\ast\!:H^\ast(\mathfrak{g})\!\otimes\!H^\ast(\mathfrak{g})\!=\!H^\ast(\mathfrak{g}\!\times\!\mathfrak{g})\longrightarrow H^\ast(\mathfrak{g})~~~\text{ and }~~~\iota^\ast\!:R\!=\!H^\ast(0)\!\longrightarrow\!H^\ast(\mathfrak{g})$$ that make $H^\ast(\mathfrak{g})$ and associative unital graded-commutative $R$-algebra, right?
What worries me here is that there is no mention of Lie brackets $[-,-]$ in this multiplication: they only appear in the boundaries of the cochain complex.
Is there a nice example of Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ such that $\mathfrak{g}\!\cong\!\mathfrak{h}$ and $H^\ast(\mathfrak{g})\!\cong\!H^\ast(\mathfrak{h})$ as modules, but $H^\ast(\mathfrak{g})\!\ncong\!H^\ast(\mathfrak{h})$ as algebras?
See this post for an explicit formula for the diagonal map.