I'm not sure you will count this as a complete answer, but here is a (stable) homotopy theoretic approach to the problem. Recall that by Brown representability, for a CW complex $X$ and an abelian group $A$ there are isomorphisms of groups
$$H_n(X;A)\cong\pi_n^S(X\wedge HA)=\{S^n,X\wedge HA\},\qquad H^n(X;A)\cong \{X,\Sigma^nHA\}.$$
Here $HA$ is the Eilenberg-Mac Lane ring spectrum associated to the abelian group $A$, with multiplication $\mu:HA\wedge HA\rightarrow HA$. The curly brackets $\{S^n,X\wedge HA\}$ represent homotopy classes of stable maps in the category of spectra, and I am supressing the (infinite) suspension functor from notation (so, for instance, I should write $\Sigma^\infty X$ for the image of $X$ in the category of spectra, but do not for readability).
If $X$ is not a CW complex, then a suitably CW replacement $\bar X$ should be taken, which should then replace $X$ in all of the previous isomorphisms.
Now the cup product of $\alpha\in H^m(X;A)$ and $\beta\in H^n(X;A)$ is the class $\alpha\cup\beta\in H^{n+m}(X;A)$ represented by the composition
$$\alpha\cup\beta:X\xrightarrow{\Delta} X\wedge X\xrightarrow{\alpha\cup\beta} \Sigma^mHA\wedge \Sigma^mHA\xrightarrow{\cong}\Sigma^{m+n}HA\wedge HA\xrightarrow{\Sigma^{m+n}\mu} \Sigma^{m+n}HA.$$
On the other hand, given $\alpha\in H^m(X;A)$ and $x\in H_k(X;A)$, their cap product is represented by the composition
$$\alpha\cap x:S^{k-m}\cong S^{-m}\wedge S^k\xrightarrow{1\wedge x} S^{-m}\wedge X\wedge HA\xrightarrow{1\wedge\Delta\wedge 1} S^{-m}\wedge X\wedge X\wedge HA\xrightarrow{1\wedge 1\wedge\alpha\wedge 1} S^{-m}\wedge X\wedge \Sigma^mHA\wedge HA\cong X\wedge HA\wedge HA\xrightarrow{1\wedge\mu} X\wedge HA,$$
and is a class in $\{S^{k-m},X\wedge HA\}\cong H_{n-k}(X;A)$. Note that the negative dimensional spheres make sense in the category of spectra.
Now if $\beta\in H^n(X;A)$ is another cohomology class, consider the diagram
$\require{AMScd}$
\begin{CD}
S^{k-m-n}@>1\wedge x>>S^{-(n+m)}\wedge X\wedge HA @>1\wedge\Delta\wedge 1>>S^{-(n+m)}\wedge X\wedge X\wedge HA@>1\wedge 1\wedge\alpha\cup \beta\wedge 1>>X\wedge HA\wedge HA\\
@VV 1\wedge 1\wedge xV @VV 1\wedge \Delta\wedge 1 V @V V 1\wedge 1\wedge \Delta\wedge 1V @VV1\wedge\mu V\\
S^{-n}\wedge S^{-m}\wedge X\wedge HA @>1\wedge1\wedge\Delta\wedge 1>> S^{-(n+m)}\wedge X\wedge X\wedge HA@>1\wedge\Delta\wedge 1\wedge 1>>S^{-(n+m)}\wedge X\wedge X\wedge X\wedge HA@>1\wedge\mu(\alpha\wedge\beta)\wedge 1>> X\wedge HA\\
@VV 1\wedge 1\wedge \Delta \wedge 1V@VV1\wedge \Delta\wedge\beta\wedge 1V@VV1\wedge\alpha\wedge \beta\wedge 1V@VV=V\\
S^{-n}\wedge S^{-m}\wedge X\wedge X\wedge HA@>1\wedge1\wedge\Delta\wedge \beta\wedge1>>S^{-n}\wedge X\wedge X\wedge HA\wedge HA@>1\wedge1\wedge\alpha\wedge 1>>X\wedge HA\wedge HA\wedge HA@>1\wedge\mu\wedge1 >>X\wedge HA
\end{CD}
When we use the equalities $\mu(1\wedge \mu)=\mu(\mu\wedge 1)$ and $(1\wedge \Delta)\Delta=(\Delta\wedge 1)\Delta$ (these are equalities of stables homotopy classes) we see that the diagram commutes in the homotopy category of spectra. Note that the anti-clockwise composition around the diagram from top-left to bottom-right represents $(\alpha\cup\beta)\cap x$, whilst the clockwise composition around the diagram represents $\alpha\cap(\beta\cap x)$. Since the diagram commutes, we see that
$$(\alpha\cup\beta)\cap x=\alpha\cap(\beta\cap x)$$
as required. Moreover, not one Eilenberg-Zilber nor Alexander-Whitney map was ever even thought about.
Edit: The big diagram might be a bit obscured, but I'm not sure how else I could format the thing to make it fit.
Best Answer
Let $X$ be a homotopy associative $H$-space with multiplication $\mu$. All homology and cohomology is taken with coefficients in a fixed field. I was unsure in the comments if it was the associativity of the homology cross product that you were unsure of, or something else. I suggested approaching the dual problem of coassociativity of the coproduct in cohomology since I know that Hatcher proves that cup products are associtive (and these are the same as the cohomology cross product). Anyway, I'll use homology in this post, and if I've misunderstood the content of your question please let me know.
I'll write $m$ for the Pontryagin product, which is the composition $$m:H_*X\otimes H_*X\xrightarrow\times H_*(X\times X)\xrightarrow{\mu_*}H_*X$$ where $\times$ is the homology cross product and $\mu:X\times X$ is the homotopy associative H-space multiplication. Then directly from the definitions we have a commutative diagram of abelian groups and homomorphisms $\require{AMScd}$ \begin{CD} H_*X\otimes H_*X\otimes H_*X@>1\otimes m>> H_*X\otimes H_*X@>m>> H_*X\\ @V\cong V 1\otimes\times V @V= V V&@V= V V\\ H_*X\otimes H_*(X\times X) @>1\otimes\mu_*>> H_*X\otimes H_*X@>m>>H_*X\\ @V\cong V\times V @V\cong V\times V&@V= V V\\ H_*(X\times X\times X) @>(1\times\mu)_*>> H_*(X\times X)@>\mu_*>>H_*X. \end{CD} The only place that commuativity is not obvious is the bottom-left square, and here it follows from the naturality of the homology cross product, a statement of which can be found as Lemma 3B.2 (pg. 270).
Now we know that homotopic maps induce the same maps on homology, and as we have assumed the existence of an associating homotopy
$$\mu\circ(1\times\mu)\simeq\mu\circ(\mu\times 1)$$
we can use functorality of the induced homomorphisms to get
$$\mu_*(1\times\mu)_*=(\mu\circ(1\times\mu))_*=(\mu\circ(\mu\times1))_*=\mu_*(\mu\times1)_*$$
for the maps on the bottom row of our diagram.
Therefore if we return to our first diagram and change the bracketing to get a second commuative diagram $\require{AMScd}$ \begin{CD} H_*X\otimes H_*X\otimes H_*X@>m\times 1>> H_*X\otimes H_*X@>m>> H_*X\\ @V\cong V \times\otimes 1 V @V= V V&@V= V V\\ H_*(X\times X)\otimes H_*X @>\mu_*\otimes 1>> H_*X\otimes H_*X@>m>>H_*X\\ @V\cong V\times V @V\cong V\times V&@V= V V\\ H_*(X\times X\times X) @>(\mu\times 1)_*>> H_*(X\times X)@>\mu_*>>H_*X \end{CD} then we find that we can paste this new diagram and our first diagram together along their bottom rows using the commutative square $\require{AMScd}$ \begin{CD} H_*(X\times X\times X)@>(1\times\mu)_*>> H_*(X\times X)\\ @VV(\mu\times_1)_* V @VV \mu_* V\\ H_*(X\times X) @>\mu_*>> H_*(X). \end{CD}
At this stage we're flying so far above my AMScd skills that I'm not going to attempt any further diagrams, but hopefully I've done enough to convince you that the result is true.