I'm not sure what 'global cohomology class' means, but for the Leray-Hirsch theorem I know of it's enough to ask that the inclusion of the fiber $G \rightarrow G \times B$ induces a surjection $H_{dR}^*(G \times B) \rightarrow H^*(G)$.
But this is clear since $G \rightarrow G \times B \rightarrow G$ is the identity, and so $G$ is a retract of $G \times B$. Since $H_{dR}^*$ is functorial it takes retracts to retracts (but flipping arrows), and so the hypotheses of the Leray-Hirsch theorem are satisfied, whence we have $H_{dR}^*(G) \otimes H_{dR}^*(B) \cong H_{dR}^*(G \times B)$.
The idea is arguably the main theme of the first half of the book, "induction on a good cover."
For $U$ a contractible open set, the restricted bundle $E|_U$ over $U$ is homeomorphic to $U \times F$, so its cohomology is just
$$H^*(E|_U) = H^*(F) = H^*(F) \otimes H^*(U),$$
since $H^*(U)$ is trivial, and has a basis given by the $e_k$.
Now consider what happens on a union $U \cup V$ of two contractible open sets?
(We will write $H^*(F) = \mathbb{R}\{e_k\}$ for a vector space, not a ring; the maps involved will typically destroy the multiplicative structure of $H^*(F)$.)
The Mayer–Vietoris sequence yields a triangle
$$H^*(E|_{U \cup V}) \to H^*(E|_U) \times H^*(E|_V) \to H^*(E|_{U \cap V}) \to \cdots,$$
with a graded vector space map from the sequence
$$H^*(F) \otimes H^*({U \cup V}) \to (H^*(F) \otimes H^*(U)) \times (H^*(F) \otimes H^*(V)) \to H^*(F) \otimes H^*({U \cap V}) \to \cdots,$$
given on the basespace factor by $\pi^*$ and on $H^*(F)$ by restriction of the global classes $e_k$ to the appropriate set. This is as in the proof of the Künneth theorem, with the difference that for that theorem, the projection $\rho\colon E = M \times F \to F$ induces the ring map $\rho^* \colon H^*(F) \to H^*(E)$ that we needed; here, although $\rho$ no longer exists, the hypothesis is that $\rho^*$ still exists, at least as a map of vector spaces.
Making this allowance, the map of sequences exists, so because we know the isomorphism for contractible sets, the five lemma gives us a $H^*(U \cup V)$-module isomorphism $$H^*(F) \otimes H^*({U \cup V}) \to H^*(E|_{U \cup V}).$$
Replacing $U$ with $U_1 \cup \cdots \cup U_{n-1}$ and $V$ with $U_n$, this is the induction step, just as in the other proof.
The only further thing I believe you need to check is the commutativity in this case of the analogue of the bottom square from the previous page.
Best Answer
Since you quote Bott--Tu, I'll use that:
They define on p.56 the pullback of a vector bundle. More generally you can define in the same manner the pullback of a fiber bundle $E$ on a space $B$ with fiber $F$ and structure group $G$: given a map $f : B' \to B$, there exists a new fiber bundle $f^{-1}E$ on $B'$ with again fiber $F$ and structure group $G$.
Theorem 6.8 (p.57) extends to this setting: if $f \sim g$ ($f$ is homotope to $g$), then $f^{-1}E \cong g^{-1}E$ (the proof is essentially the same).
It is also true that $(g \circ f)^{-1}E \cong f^{-1}(g^{-1}E)$, just like for vector bundles.
Now if $B$ is a Euclidean space, it's contractible, meaning there are map $f : B \to \{*\}$ and $g : \{*\} \to B$ such that $f \circ g \sim id_{\{*\}}$ and $g \circ f \sim id_B$. Now if you apply that last equality to any fiber bundle $E$ on $B$, you get $f^{-1}(g^{-1}E) \cong id^{-1}E = E$. But a fiber bundle on a singleton is necessarily trivial (directly from the definitions), thus $g^{-1}E$ is trivial. And the pullback of a trivial bundle is trivial again, so $E = f^{-1}(g^{-1}E)$ is trivial again.