I want to show:
If $\xi$ is isomorphic to $\eta$ then $w_i(\xi)=w_i(\eta)$
Does isomorphic here mean isomorphic bundles over the same base space? I.e. if I have an isomorphism then:
I know that I have $\psi$ is a homeomorphism, and that $\pi_1^{-1}(b)\cong \pi_2^{-1}(b)$ is an isomorphism for all $b$. Firstly, it seems then I have that $\pi_1^{-1}(b),\pi_2^{-1}(b)$ and hence that these have the same base space (i.e. $b\in B(\xi),b\in B(\eta)$). \begin{align} \begin{matrix} E(\xi)&\stackrel{\psi}\longrightarrow&E(\eta)\\ \!\!\!\!\!\!\!\pi_1\downarrow&& \quad\downarrow\pi_2\\ B(\xi)&\dashrightarrow&B(\eta) \end{matrix} \end{align}
Anyway, with that in mind, $\pi_2(\psi(\pi_1^{-1}(b)))=\pi_2(\pi_2^{-1}(b))=b$ tells us that $f:B(\xi)\to B(\eta)$ is the identity map, relying on $\pi_1^{-1}(b)\cong \pi_2^{-1}(b)$. Then the induced homomorphism is the identity homomorphism on the cohomology and we are done by naturality.
Yes, we normally talk about "isomorphism of bundles" only for bundles on a fixed base space. This means that in the diagram you've written, $B(\xi)=B(\eta)$ and the diagram should commute with the bottom map being the identity map.