Supposed I do not define ordered pair in the usual Kuratowski way $(x,y) = \{\{x\},\{x,y\}\}$. I left the ordered pair undefined but with the propriety $(x,y) = (x',y') \iff x=x'\text{ and }y=y'$. Then I postulate the axiom:
For every sets $A$, $B$, there exist one unique set $\sf CP$ (Cartesian Product) such that $(x,y)\in\mathsf{CP}\iff x\in A\text{ and }y\in B$.
The remain $\sf ZFC$ axioms unchanged. Can be a contradiction be found with under this conditions?
Well. What is $(x,y)$?
In the language of set theory there is only a binary relation symbol $\in$, which $x\in y$ is interpreted to mean "$x$ is an element of $y$".
If you added a new function symbol $O(x,y)$ and the axioms that $O(x,y)=O(x',y')$ if and only if $x=x'$ and $y=y'$, then you could argue that by adding the axiom that for every $A$ and $B$ there is a set $\sf CP(A,B)$ which is effectively $A\times B$, you didn't change anything.
But you still need to address something. Now that you have added a new function symbol to your language, you have more formulas. So you need to decide whether or not you want to extend the Replacement schema or not, or other schemata if you have them.
In general, however, this is easily equiconsistent with $\sf ZFC$, since it is in fact a conservative extension. We have a definable $O(x,y)$ function, namely the Kuratowski (or any other) definition of an ordered pair.
But the beauty of set theory is in its simplicity. The fact that a few axioms (and a schema) can do so much with just one binary relation symbol.