If you intend $x,y,z$ to be naturals, you can use the Cantor pairing function, which allows you to recover both $x$ and $y$ from $z$, though the functions are different. That is, you have $z=f(x,y)$ and there is a function $g(z)$ which returns the unique $x$ which could have given that $z$ and another function $h(z)$ which returns the $y$. You could say $(g(z),h(z))=f^{-1}(z)$ but as $f$ takes an ordered pair as input, the output of $f^{-1}$ has to be an ordered pair, not a single number as in your question. It is not monotonic in the way you want, as $f(2,1)$ would have to have an infinite number of predecessors, all the $f(1,k)$ and no natural does.
I think you have the same problem over the reals. To get the monoticity you ask, you would have to have f(x,y) occupy an interval. Say $f(0,y)$ ranges from $0$ to $1$. Then $f(1,y')$ would have to be greater than $1$. But you can't have uncountably many intervals in the real line.
It seems possible to me that you could do this in $\mathbb Q$ because the order types are compatible, but I haven't figured it out yet. Vladimir Reshetnikov has been working on this lately.
Added: It can be done in $\mathbb Q$. My construction shows existence, but does not result in an explicit function. Perhaps by being more explicit in the construction the function can be made explicit. Basically we follow the proof that any two countable dense linear orders without endpoints are isomorphic. List the rationals in order as $r_1,r_2,r_3,\ldots$. We will assign $f(r_i,0)$ in such a way that we can fit $f(r_i,y)$ into an interval disjoint from all the other intervals, maintaining the order as desired. So let $f(r_1,0)=0$ Assign it the interval $(-\frac 12,\frac 12)$, so $f(r_1,y) \in (-\frac 12,\frac 12)$ and the order is maintained. Then assign $f(r_2)$ to $\pm 2$, whichever is required to keep it in order relative to $r_1$. Assign it the interval$(-\frac 52,-\frac 32)$ or $(\frac 32,\frac 52)$ as appropriate and put all the $f(r_2,y)$ into that interval in proper order. Continue assigning $f(r_i,0)$ in order, putting them in proper order with respect to the earlier $f(r_j,0)$ and giving it a small interval around $f(r_i,0)$ to put the $f(r_i,y)$. By shrinking the intervals enough, they will never overlap, so the desired order is maintained.