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,η,∂) (see here, chapter XV.7) consists of modules H(p,q) for each p≤q, morphisms η:H(p′,q′)→H(p,q) for all p≤p′, q≤q′, and boundary morphisms ∂:H(p,q)→H(q,r) for all p≤q≤r, such that
η=id:H(p,q)→H(p,q),
η=η∘η:H(p″,
\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 Er will relate in any sensible way to the differential of Er+1! So you can construct examples by just picking a product on E3, say, which does not do what you want it not to do... – Mariano Suárez-Álvarez May 09 '10 at 15:16