If $f(u(x), v(y))=f(x, y)$, can we conclude that either $u(x)=x$ or $v(y)=y$?

The answer is no. Take $f(x,y)=x^2-1+y^2$ and $u(x)=-x,$ $\nu(y)=-y.$ If you want some more "nontrivial" examples, you can consider symmetries with respect to $x\to 1-x$ or something like that (instead of obvious one $x\to -x.$)

You can have $f(x,y)=x+y$, and then $u$ can shift by a constant, with $v$ shifting in reverse.