I realized that your question wasn't exactly about the Steenrod axioms themselves, but about the definition of the coboundary operator involving suspension. In reduced homology, the boundary operator $\partial$ for the pair $(X,A)$ (where the inclusion $i:A\rightarrow X$ is a cofibration) can be defined to come from the "topological boundary map" $\partial^!$ followed by the inverse of the suspension isomorphism. The former is itself a composition
$$ \partial^! = \pi \circ \psi^{-1}: X/A \rightarrow Ci \rightarrow \Sigma A, $$
where $Ci$ is the mapping cone of $i$, $\psi^{-1}$ is a homotopy inverse of the quotient $\psi: Ci \rightarrow Ci/CA=X/A$, and $\pi: Ci \rightarrow Ci/X=\Sigma A$. So
$$ \partial = (\Sigma_*)^{-1} \circ \partial^!_* : \tilde{H}_q(X/A) \rightarrow \tilde{H}_q(\Sigma A) \rightarrow \tilde{H}_{q-1}(A) .$$
In fact, this is true for any reduced homology theory. See May's "Concise Course" for details, pp. 106-7. I'm pretty sure that the situation for cohomology is very similar.
Bottom line: In this formulation, the coboundary operator is the composition of a map induced from an actual map on spaces and the (inverse of the (?)) suspension isomorphism. Steenrod squares commute with both of these, so they commute with the coboundary operator.