TITLE: Generalizing detropicalization
QUESTION [9 upvotes]: Given an identity in max,plus arithmetic, are there ways to turn it into an ordinary algebraic identity it other than by replacing addition by multiplication and replacing max by series-plus or by parallel-plus, where the series sum of $x$ and $y$ is $x+y$ and the parallel sum is $xy/(x+y)$?  (See the related posts choosing between the two ways to tropicalize and Name and notation for a binary operation.)
As a related question, I ask: What are all the two-variable homogeneous rational functions over $\mathbb{C}$ such that $r(x,y)=r(y,x)$ and $r(r(x,y),z)=r(x,r(y,z))$?  (Here I intend equality in the formal sense; e.g., I call $x/x$ the same rational function as 1, even though as functions they are not the same, since $x/x$ is not defined at 0.)
The only examples I know of are $r(x,y) = c$ (with $c$ an arbitrary constant), $r(x,y)=x+y$, $r(x,y)=xy/(x+y)$, and $r(x,y)=xy$, but I suspect that there are others I'm overlooking.  Perhaps the theory of formal groups has an answer for me, but from what little I've seen, homogeneity does not play a role there.  See the related post
Commutative associative rational binary operations.  Now that I've played a bit with operations like $(x+y)/(1-xy)$ (thanks, Alexandre Eremenko!), I'm suspecting that I should be limiting myself to homogeneous operations.
(Note: The original version of the post used the word "detropicalize'' in a non-standard and unclear way, so I've changed the title of the post and the wording of the first paragraph accordingly.  Thanks, John Mangual!)

REPLY [6 votes]: Those are all.
Given a function $r$, by restricting to a particular value of $y$ (barring finitely many), we get a rational function, hence a map $\mathbb P^1 \to \mathbb P^1$. For all but finitely many values of $y$, this map will have the same degree, $d$. Assuming $r$ is nonconstant, let $y_1$ and $y_2$ be two such typical values such that $r(y_1,y_2)$ is also typical. Then the degree of $x => r(x,y_1) => r(r(x,y_1),y_2)$ is the degree of $x => r(x,r(y_1,y_2))$, so $d^2=d$,so $d=0$ or $1$. Clearly the $d=0$ case is the constant case.
For $d=1$, we get a rational inverse function to $r$, giving us an algebraic group structure on $\mathbb P^1$ minus finitely many points. (Or for each $y$-value but finitely many, we get an automorphism of $\mathbb P^1$, which is an element of $PGL_2$, so we have a curve in $PGL_2$ that is almost closed under composition. The closure of such a curve is always a subgroup.) There are only two of these: $\mathbb G_a$ or $\mathbb G_m$. But we still have to decide which isomorphism between the closure of these groups and $\mathbb P^1$ to take. The homogeneity means that the missing points can only be $0$ and $\infty$, which gives us three options:
$\mathbb G_a$, $\infty$ missing: $r(x,y)=x+y$
$\mathbb G_a$, $0$ missing: $r(x,y) = 1/((1/x)+(1/y))=xy/(x+y)$
$\mathbb G_m$, both $0$ and $\infty$ missing: $r(x,y)= xy$.