I am sorry that I don't know a reference, so I will try to give a proof instead.
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 $(\tilde h_\bullet,\partial,\wedge)$ be a reduced multiplicative generalised homology theory, and let $(\tilde h^\bullet,\delta,\wedge)$ be the corresponding cohomology theory. We define Cartan-Eilenberg systems $(H,\eta,\partial)$, $(H',\eta',\partial')$ by
$$H(p,q)=\tilde h_\bullet(X^{-p},X^{-q})\qquad\text{and}\qquad
H'(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$
for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$, $\eta'\colon H'(p',q')\to H'(p,q)$ for $p\le p'$, $q\le q'$.
The maps $\partial\colon H(p,q)\to H(q,r)$, $\partial'\colon H'(p,q)\to H'(q,r)$ come from exact sequences of triples.
Sorry for the weird indexing,
but this makes it the easiest to use Douady's result.
The spectral sequence will not end up in the first quadrant, but you can surely bring it back into its rightful place.
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}/(X\wedge X)^{m+n-r}
\cong\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/
\bigcup_{c+d=m+n-r}(X^c\wedge X^d)\\
\begin{aligned}
\twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/
\Bigl(\bigcup_{a=0}^{m-1}(X^a\wedge X^{m+n-a})
\cup\bigcup_{b=0}^{n-r}(X^{m+n-b}\wedge X^b)\\
\cong\mathord{}&\bigcup_{a=0}^{r-1}(X^{m+a}\wedge X^{n-a})\Bigm/
\bigl(X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r}\bigr)\\
\hookrightarrow\mathord{}& X^{m+r-1}\wedge X^n\bigm/
(X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r})\\
\cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r})\;.
\end{aligned}
\end{multline*}
Together with the diagonal map $\Delta$, for $r\ge 1$, we define
\begin{multline*}
\mu_r\colon H(-n-m,r-n-m)\otimes H'(m,m+r)\\
\begin{aligned}
&\cong\tilde h_\bullet(X^{m+n}/X^{m+n-r})\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
&\stackrel{\Delta_X^*\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet\bigl((X\wedge X)^{m+n}/(X\wedge X)^{m+n-r}\bigr)\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
&\stackrel{F_{m,n,r,*}\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet((X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r}))\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
&\stackrel{/}\longrightarrow\tilde h_\bullet(X^{n}/X^{n-r})=H(-n,r-n)\;.
\end{aligned}
\end{multline*}

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

As explained here, this Proposition allows us to define a multiplicative structure on the associated spectral sequence. In this setting, it will be the cap product you ask for:
$$\frown\colon E^r_{m+n}\otimes E_r^m\to E^r_n\;.$$

**Proof.**
The upper square commutes because the maps $F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we use the Leibniz rule for the slant product and continue as here