We follow Douady's approach using Cartan-Eilenberg systems, see here.

Let $B$ be a CW complex and $\pi\colon X\to B$ a Serre fibration. Put $X^k=\pi^{-1}(B^k)$. A cellular approximation~$\Delta_B\colon B\to B\times B$ of the diagonal can be lifted to an approximation $\Delta\colon X\to X\times X$ of the diagonal such that
$$X^k\stackrel\Delta\longrightarrow\bigcup_{m+n=k}X^m\wedge X^n\;.$$

Let $(\tilde h^\bullet,\delta,\wedge)$ be a reduced multiplicative generalised cohomology theory. We define a Cartan-Eilenberg system $(H,\eta,\partial)$ by
$$H(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$
for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$ for $p\le p'$, $q\le q'$. The corresponding exact sequences take the form
$$\cdots\to\tilde h^\bullet(X^{r-1},X^{q-1})\to\tilde h^\bullet(X^{r-1},X^{p-1})
\to\tilde h^\bullet(X^{q-1},X^{p-1})\stackrel\delta\to
\tilde h^\bullet(X^{r-1},X^{q-1})\to\cdots$$
We ignore the grading; it is easy to fill in.

To define a spectral product $\mu\colon(H,\eta,\partial)\times(H,\eta,\partial)\to(H,\eta,\partial)$ we consider the map
\begin{multline*}
F_{m,n,r}\colon(X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1}
\cong\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/
\bigcup_{c+d=m+n-1}(X^c\wedge X^d)\\
\begin{aligned}
\twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/
\Bigl(\bigcup_{a=0}^m(X^{a-1}\wedge X^{m+n+r-a})
\cup\bigcup_{b=0}^n(X^{m+n+r-b}\wedge X^{b-1})\\
\cong\mathord{}&\bigcup_{a=m+1}^{m+r}(X^{a-1}\wedge X^{m+n+r-a})\Bigm/
\bigl(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1}\bigr)\\
\hookrightarrow\mathord{}& X^{m+r-1}\wedge X^{n+r-1}\bigm/
(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1})\\
\cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\;.
\end{aligned}
\end{multline*}Together with the diagonal map $\Delta$, for $r\ge 1$, we define
\begin{multline*}
\mu_r\colon H(m,m+r)\otimes H(n,n+r)
\cong\tilde h(X^{m+r-1}/X^{m-1})\otimes\tilde h(X^{n+r-1}/X^{n-1})\\
\begin{aligned}
&\stackrel\wedge\longrightarrow\tilde h\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
&\stackrel{F_{m,n,r}^*}\longrightarrow\tilde h\bigl((X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1}\bigr)\\
&\stackrel{\Delta_X^*}\longrightarrow\tilde h(X^{m+n+r-1}/X^{m+n-1})=H(m+n,m+n+r)\;.
\end{aligned}
\end{multline*}

**Proposition**
*For all $m$, $n$, $r\ge 1$, the following diagram commutes*
$\require{AMScd}$
\begin{CD}
H(m,m+1)\otimes H(n,n+1)@>\mu_1>>H(m+n,m+n+1)\\
@A\eta\oplus A\eta A@AA\eta A\\
H(m,m+r)\otimes H(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\
@V\partial\otimes\eta\oplus V\eta\otimes\partial V@VV\partial V\\
{\begin{matrix}H(m+r,m+r+1)\otimes H(n,n+1)\\\oplus\\H(m,m+1)\otimes H(n+r,n+r+1)\end{matrix}}@>\mu_1\pm\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{;,}
\end{CD}

As explained here, this Proposition allows us to define a multiplicative structure on the associated spectral sequence.

**Proof.**
The upper square commutes because the maps~$F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we consider the boundary morphism $\delta$ of the triple
$$(X^{m+r}\wedge X^{n+r-1}\cup X^{m+r-1}\wedge X^{n+r},
X^{m+r}\wedge X^{n-1}\cup X^{m+r-1}\wedge X^{n+r-1}\cup X^{m-1}\wedge X^{n+r},\\
X^{m+r}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r})\;.$$
The following diagram commutes:
\begin{CD}
\tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1})
@>\wedge>>
\tilde h^{-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
@V\delta\wedge\mathrm{id}\oplus V\mathrm{id}\wedge\delta V
@VV\delta V\\
{\begin{matrix}
\tilde h^{1-p}(X^{m+r}/X^{m+r-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1})\\
\oplus\\
\tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{1-q}(X^{n+r}/X^{n+r-1})
\end{matrix}}
@>\wedge\oplus\wedge>>
{\begin{matrix}
\tilde h^{1-p-q}\bigl((X^{m+r}/X^{m+r-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
\oplus\\
\tilde h^{1-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r}/X^{n+r-1})\bigr)
\end{matrix}}
\end{CD}
We extend this diagram to the right using the maps $F_{m,n,r}$ and
conclude that the lower square also commutes.

extremelytedious." $\endgroup$4more comments