You cannot extend the Steenrod squares to integer coefficients (as in have a set of cohomology operations satisfying the same axioms).

Consider the tangent space $TS^2$ of the 2-sphere. With a little persistence, we can identify the Thom space of this vector bundle with the two cell complex $S^2 \cup_{2\eta} e^4$ (here $\eta$ is as usual the Hopf map generating $\pi_3(S^2))$. This map $\eta$ has Hopf invariant 1, and so the cup product on this Thom space is $x \cup x = 2y$. Now the tangent bundle of the sphere is trivial after adding a trivial line bundle, so the suspension of this space is $S^3 \vee S^5$. By naturality, any cohomology operation applied to the 3-cell is trivial.

Hence, there is no stable transformation $Sq^2:H^*(-) \rightarrow H^{*+2} (-)$ so that for a 2 dimensional cohomology class x, $Sq^2 (x)=x \cup x$.

You cannot extend the Steenrod squares to integer coefficients (as in have a set of cohomology operations satisfying the same axioms).

Consider the tangent space $TS^2$ of the 2-sphere. With a little persistence, we can identify the Thom space of this vector bundle with the two cell complex $S^2 \cup_{2\eta} e^4$ (here $\eta$ is as usual the Hopf map generating $\pi_3(S^2))$. This map $\eta$ has Hopf invariant 1, and so the cup product on this Thom space is $x \cup x = 2y$. Now the tangent bundle of the sphere is trivial after adding a trivial line bundle, so the suspension of this space is $S^3 \vee S^5$. By naturality, any cohomology operation applied to the 3-cell is trivial.

Hence, there is no stable transformation $Sq^2:H^*(-) \rightarrow H^{*+2} (-)$ so that for a 2 dimensional cohomology class x, $Sq^2 (x)=x \cup x$.

Of course, there other ways to see that the suspension of $2 \eta$ is trivial, but I like this one.