The Leray–Hirsch theorem says that given a fiber bundle $F \to E \to B$ such that $H^*(F)$ is free (as a module over whatever coefficient ring $k$) and, for each $n \geq 0$ there is a set of classes $H^n(E)$ whose restrictions form a basis for the cohomology of each fiber, we have an $H^*(B)$-module isomorphism
$$H^*(E) \cong H^*(B) \otimes H^*(F).$$
It is further the case that if $k$ is a field of characteristic zero (I think any field will do) and $E \to B$ is a principal $G$-bundle for some compact, connected Lie group $G$, then the isomorphism
$$H^*(E) \cong H^*(B) \otimes H^*(G)$$
can in fact be taken to be a ring isomorphism. This can be seen as a Künneth theorem, even though the bundle is not a product.
The proof that I know involves connections and is scattered through a few chapters of lemmas in a fairly lengthy book. I feel that (okay, would like it to be the case that) the action $E \times G \to G$, restricting on each fiber to the multiplication of $G$, and the fact that $H^*(G)$ is naturally a Hopf algebra should together lead to some sort of diagram-theoretic proof of this Künneth theorem.
So, is there some obvious reason for this ring isomorphism?
This is true for the dumbest of possible reasons. If $G$ is a compact, connected Lie group, then after inverting finitely many primes in the coefficient ring, $H^*(G) = \Lambda P$ is an exterior algebra.
Over characteristics other than 2, exterior algebras are free commutative graded algebras (CGAs), so any surjection $H^*(E) \to H^*(G)$ splits. One simply lifts a basis for $P$ up to $H^*(E)$ and notes that these elements span an exterior subalgebra of $H^*(E)$ because it is a CGA. This isomorphic subalgebra then lets one define a CGA map $H^*(B) \otimes H^*(G) \to H^*(E)$ which commutes with the maps from $H^*(B)$ and to/from $H^*(G)$. At least if each $H^j(B)$ is of finite rank, this map is bijective simply by rank considerations.