This is an expansion of John Rognes' answer. I have filled in a few details in Douady's seminare notes and noticed that one gets away with slightly weaker axioms. If there is already a reliable reference for all this, please let me know.
Recall that a Cartan-Eilenberg system $(H,\eta,\partial)$ (see here, chapter XV.7) consists of modules $H(p,q)$ for each $p\le q$, morphisms $\eta\colon H(p',q')\to H(p,q)$ for all $p\le p'$, $q\le q'$, and boundary morphisms $\partial\colon H(p,q) \to H(q,r)$ for all $p\le q\le r$, such that
$\eta=\mathrm{id}\colon H(p,q)\to H(p,q)$,
$\eta=\eta\circ\eta\colon H(p'',q'')\to H(p',q')\to H(p,q)$,
$\eta$ and $\partial$ commute,
there are long exact sequences $\cdots\to H(q,r)\stackrel\eta\to H(p,r)\stackrel\eta\to H(p,q)\stackrel\partial\to H(q,r)\to\cdots$.
The conditions needed for convergence have been omitted. A typical example is $H(p,q)=\tilde h_\bullet(X_p/X_q)$, where $\cdots\supset X_{-1}\supset X_0\supset X_1\supset\cdots$ is a decreasing sequence of cofibrations and $\tilde h_\bullet$ is some generalised homology theory. The grading is suppressed in the following, but you can easily fill it in.
To set up a spectral sequence from $(H,\eta,\partial)$, one defines
$$Z^r_p=\mathrm{im}\bigl(H(p,p+r)\stackrel\eta\to H(p,p+1)\bigr)\;,$$
$$B^r_p=\mathrm{im}\bigl(H(p-r+1,p)\stackrel\partial\to H(p,p+1)\bigr)\;,$$
$$E^r_p=Z^r_p/B^r_p\;,$$
$$d^r_p\colon Z^r_p/B^r_p\twoheadrightarrow Z^r_p/Z^{r+1}_p\cong
B^{r+1}_{p+r}/B^r_{p+r}\hookrightarrow Z^r_{p+r}/B^r_{p+r}\;.$$
Details are in Switzer's book, chapter 15. In particular
$$\ker(d^r_p)=Z^{r+1}_p/B^r_p\qquad\text{and}\qquad
\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p\;.$$
For $a=\eta(a_0)\in H(p,p+1)$, $a_0\in H(p,p+r)$, one has
$$d^r_p([a])=[\partial a_0]\in E^r_p\qquad\text{with}\qquad\partial a_0\in H(p+r,p+r+1)\;.$$
Definition (Douady, II.A)
Let $(H,\eta,\partial)$, $(H',\eta',\partial')$ und $(H'',\eta'',\partial'')$ be Cartan-Eilenberg systems. A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ is a sequence of maps
$$\mu_r\colon H'(m,m+r)\otimes H''(n,n+r)\to H(m+n,m+n+r)$$
such that for all $m$, $n$, $r\ge 1$, the following two diagrams commute.
$\require{AMScd}$
\begin{CD}
H'(m,m+r)\otimes H''(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\
@V\eta'\oplus V\eta''V@VV\eta V\\
H'(m,m+1)\otimes H''(n,n+1)@>\mu_1>>H(m+n,m+n+1)\rlap{\;,}
\end{CD}
\begin{CD}
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+\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{\;.}
\end{CD}
The first diagram is weaker than in Douady's notes.
The second can be read as a Leibniz rule.
Theorem (Douady, Thm II)
A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ induces products
$$\mu^r\colon E^{\prime r}_m\otimes E^{\prime\prime r}_n\to E^r_{m+n}\;,$$
such that
$\mu^1=\mu_1$
$d^r_{m+n}\circ\mu^r=\mu^r\circ(d^{\prime r}_m\otimes\mathrm{id})\pm\mu^r\circ(\mathrm{id}\circ d^{\prime\prime r}_n)$,
$\mu^{r+1}$ is induced by $\mu^r$.
Proof.
Assume by induction that $\mu^r$ is induced by $\mu_1$. In particular,
$$Z^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to Z^r_{m+n}\;,$$
$$B^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;,$$
$$Z^{\prime r}_m\otimes B^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;.$$
This is clear for $r=1$ if we put $\mu^1=\mu_1$ because $E^1_p=Z^1_p=H(p,p+1)$
and $B^1_p=0$.
Let $[a]\in Z^{\prime r}_m$, $[b]\in Z^{\prime\prime r}_n$ be represented by $a=\eta'(a_0)\in H'(m,m+1)$, $b=\eta''(b_0)\in H''(n,n+1)$ with $a_0\in H'(m,m+r)$, $b_0\in H''(n,n+r)$.
Using the first diagram and the construction of $d^r_{m+n}$, we conclude that
$$(d^r_{m+n}\circ\mu^r)([a]\otimes[b])=d^r_{m+n}[\mu_1(a\otimes b)]=d^r_{m+n}[\eta(\mu_r(a_0\otimes b_0))]=(\partial\circ\mu_r)(a_0\otimes b_0)\;.$$
From the second diagram, we get
$$(\partial\circ\mu_r)(a_0\otimes b_0)=\mu_1(\partial'a_0\otimes\eta''b_0)\pm\mu_1(\eta'a_0\otimes\partial''b_0)=\mu^r(d^{\prime r}_m[a]\otimes[b])\pm\mu^r([a]\otimes d^{\prime\prime r}_n[b])\;.$$
This proves the Leibniz rule (2).
From the Leipniz rule and the facts that $\ker(d^r_p)=Z^{r+1}_p/B^r_p$ and $\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p$, we conclude that $\mu^r$ induces a product on $E^{r+1}_p\cong\ker(d^r_p)/\mathrm{im}(d^r_p)$, which proves (3). Because $\mu^r$ is induced by $\mu_1$, so is $\mu^{r+1}$, and we can continue the induction.
$E^{r+1}=H(E^r)$by the one you defined on $E^r$ will relate in any sensible way to the differential of $E^{r+1}$! So you can construct examples by just picking a product on $E^3$, say, which does not do what you want it not to do... $\endgroup$