You can give a proof of multiplicativity by using that the smash product preserves connectivity. Here is a sketch proof.

(EDIT: Denis Nardin pointed me towards [this reference by Dugger][1]. This reference shows that I was a little too cavalier about pairings in the homotopy category vs. lifting them to the stable category, and I've tried to make some adjustments accordingly.)

For any $n$, let $\tau_{\geq n} E$ be the $(n-1)$-connected cover of $E$, assembling into the Whitehead tower
$$
\dots \to \tau_{\geq 2} E \to \tau_{\geq 1} E \to \tau_{\geq 0} E \to \tau_{\geq -1} E \to \dots \to E.
$$
The associated graded consists of the spectra $\tau_{\geq n} E / \tau_{\geq n+1} E = K(\pi_n E, n)$.

The smash product of an $(n-1)$-connected object with an $(m-1)$-connected object is $(n+m-1)$-connected, and so the composite maps
$$
\tau_{\geq n} E \wedge \tau_{\geq m} E \to E \wedge E \to E
$$
lift, essentially uniquely up to homotopy, to $\tau_{\geq n+m} E$. This means that, in the homotopy category, the tower $\{\tau_{\geq n} E\}$ forms a _filtered algebra_. Because these lifts are essentially unique, this multiplication can also be lifted in a more structured one, lifting the Whitehead tower to a filtered algebra in spectra. (To be precise, a filtered algebra is a lax symmetric monoidal functor from the symmetric monoidal poset $(\Bbb Z, \geq, +)$ to spectra.)

The smash product is symmetric monoidal. Therefore, if $F$ is a ring spectrum the tower $\{(\tau_{\geq n} E) \wedge F\}$ is also a filtered algebra, and the associated graded consists of the spectra $K(\pi_n E, n) \wedge F$ because the smash product preserves cofibers. Therefore, the associated-graded spectral sequence starts with the homology of $F$ with coefficients in the homotopy groups of $E$: this constructs the Atiyah-Hirzebruch spectral sequence, except that the indexing is slightly different (the $E_1$-term is a reindexed $E_2$-term of the Atiyah-Hirzebruch SS).

Thus, this boils down to the following assertion:

>If $\{\dots \to R^{-1} \to R^0 \to R^1 \to \dots\}$ is a (commutative / associative / unital) filtered algebra, then the associated-graded spectral sequence with
$$
E^1_{p,q} = \pi_{p+q}(R^p / R^{p-1})
$$
is multiplicative (and commutative / associative / unital).

This is a little more standard. Dugger's reference that I linked to above gives a proof; he also says that he "found the existing literature extremely frustrating," which I don't think will surprise many people.

  [1]: https://arxiv.org/abs/math/0305173