Principal fibrations with a section are trivial

I'm trying to prove that, given a principal fibration $\Omega B \rightarrow F \stackrel{p}{\rightarrow} E$ such that $p$ is a retraction of $F$ onto $E$, the total space $F$ is homotopy equivalent to the product $\Omega B \times E$. This is essentialy a problem from Hatcher (4.3, Q22).

Anyway, since $E$ is a retract of $F$, we have a map $s\colon E \rightarrow F$ such that $p\circ s = 1$. So the long exact sequence in homotopy for the fibration breaks into split short exact sequences, and one approach I've been trying is to show that the isomorphism in the splitting is induced from some map $F \rightarrow \Omega B \times E$, so that Whitehead's theorem can finish the job (you can assume I'm always talking about CW complexes).

Another approach I've tried - inspired by the proof of the similar statement about principal bundles and global cross-sections - is to try and define an action of the fibre $\Omega B$ on $F$. Up to homotopy, we can replace $F$ with the homotopy fibre $F_q$, where $E \stackrel{q}{\rightarrow} B$ is the rest of the fibration sequence in the principal fibration, and it's then very easy to define an action of $\Omega B$ on $F_q$ (again, see the exercises in Hatcher), so we have an action of $\Omega B$ on $F$ up to homotopy. So we can define a map $\Omega B \times E \rightarrow F$ as the composite $$\Omega B \times E \stackrel{1 \times s}{\rightarrow} \Omega B \times F \rightarrow F,$$ where the last map is the action, and I thought about trying to show that this induces isomorphisms in homotopy for, once again, a solution by Whitehead.

Alternatively, I could try and define a map $F \rightarrow \Omega B \times E$ and show that it's the homotopy inverse of the above map, but I've come a little unstuck in trying to do this. There's no obvious map $F \rightarrow\Omega B$.

I've been wrestling with this problem for the last few days, and I've reached the point where I'm just going around in circles and finding it hard to make much progress. Sorry for the hodgepodge nature of this post, but I wanted to explain the various ways I've been going about solving this. I'm going to leave it on the backburner for a few days, sleep on it a little more, and hopefully I'll come back to it and the answer will be obvious --- it does feel like I'm missing something important that's staring me in the face. In the meantime, though, I'd love it if someone could suggest a hint or two, or let me know if I'm tackling this in the right way!

My supervisor pointed me in the right direction for the solution, which I'm recording here for posterity. It (unsurpringly) involves the action of $\Omega B$ on the homotopy fibre $F$, just like in the bundle case.

Recall that I had defined a map $\Omega B \times E \rightarrow F$ by composition of $1 \times s$ with the action $*$. Now, remembering that the product of two fibrations is a fibration, we have the following commutative diagram of long exact sequences of homotopy groups and maps between them:

$$\begin{array}{ccccc}\pi_{n+1} E & = & \pi_{n+1} E & = & \pi_{n+1} E \\ \downarrow & & \downarrow & & \downarrow \\ \pi_{n} \Omega B & \longrightarrow & \pi_{n} \Omega B \times \pi_{n} \Omega B & \longrightarrow & \pi_n \Omega B\\ \downarrow & & \downarrow & & \downarrow \\ \pi_{n} \Omega B \times \pi_{n} E & \longrightarrow & \pi_{n} \Omega B \times \pi_n F & \longrightarrow & \pi_{n} F \\ \downarrow & & \downarrow & & \downarrow \\ \pi_{n} E & = & \pi_{n} E & = & \pi_{n} E \\ \downarrow & & \downarrow & & \downarrow \\ \pi_{n-1} \Omega B & \longrightarrow & \pi_{n-1} \Omega B \times \pi_{n-1} \Omega B & \longrightarrow & \pi_{n-1} \Omega B \end{array}$$

where the second and fifth rows come from (left) inclusion and $H$-group multiplication, and the middle row is the homomorphism induced by the map $\ast \circ (1 \times s)$ defined above. The result then follows from the 5-lemma and Whitehead's theorem.

(Sorry about the shoddiness of the diagram, I wasn't sure if this textbox is xy enabled or not, so I stuck with an array.)