3 added 138 characters in body

Rather than simply give a counterexample, I will try to explain a rationale for an emphatic "no."

So consider an operation $a\oplus b = f(g(a)+h(b))$ of your form. Now viewed as function of two variables, $a\oplus b$ will have its various level sets. Observe that changing, say, one strictly increasing $f$ for another does not change the set of level sets (one needs to say something a bit more complicated for weakly increasing $f$).

Thus, from the point of view of the geometry of the level sets, we don't lose much by taking $f={\rm id}$, the identity function. But the level sets of $g(a)+h(b)$ have very special structure: roughly from three of them you can calculate a fourth (I don't need continuity, but for simplicity I'll tacitly assume "nice" in whatever way I need) as follows:

Suppose $L \lt M,N$. I'll use the $L$,$M$ and $N$-level sets to locate points on the $M+N-L$-level set. Given $(a,b)$ such that $g(a)+h(b)=L$, find b' so that $g(a)+h(b')=M$ and a' such that $g(a')+h(b)=N$. Then $(a',b')$ sits on the desired level set. Now one can easily engineer binary operations compatible with the usual order that don't have level sets that behave this way- indeed . Indeed one may interpret the desired condition on the operation as a rather weak condition on the geometry of the level sets themselves!, namely that given a point on a level set, the level set does not pass into the quadrant to the right and above the point.

2 added 12 characters in body

Rather than simply give a counterexample, I will try to explain a rationale for an emphatic "no."

So consider an operation $a\oplus b = f(g(a)+h(b))$ of your form. Now viewed as function of two variables, $a\oplus b$ will have its various level sets. Observe that changing, say, one strictly increasing $f$ for another does not change the set of level sets (one needs to say something a bit more complicated for weakly increasing $f$).

Thus, from the point of view of the geometry of the level sets, we don't lose much by taking $f={\rm id}$, the identity function. But the level sets of $g(a)+h(b)$ have very special structure: roughly from three of them you can calculate a fourth (I don't need continuity, but for simplicity I'll tacitly assume "nice" in whatever way I need) as follows:

Suppose $L \lt M,N$. I'll use the $L$,$M$ and $N$-level sets to locate points on the $M+N-L$-level set. Given $(a,b)$ such that $g(a)+h(b)=L$, find b' so that $g(a)+h(b')=M$ and a' such that $g(a')+h(b)=N$. Then $(a',b')$ sits on the desired level set. Now one can easily engineer binary operations compatible with the usual order that don't have level sets that behave this way - indeed one may interpret the desired condition as a rather weak condition on the geometry of the level sets themselves!

1

Rather than simply give a counterexample, I will try to explain a rationale for an emphatic "no."

So consider an operation $a\oplus b = f(g(a)+h(b))$ of your form. Now viewed as function of two variables, $a\oplus b$ will have its various level sets. Observe that changing, say, one strictly increasing $f$ for another does not change the set of level sets (one needs to say something a bit more complicated for weakly increasing $f$).

Thus, from the point of view of the geometry of the level sets, we don't lose much by taking $f={\rm id}$, the identity function. But the level sets of $g(a)+h(b)$ have very special structure: roughly from three of them you can calculate a fourth (I don't need continuity, but for simplicity I'll tacitly assume "nice" in whatever way I need) as follows:

Suppose $L \lt M,N$. I'll use the $L$,$M$ and $N$-level sets to locate points on the $M+N-L$-level set. Given $(a,b)$ such that $g(a)+h(b)=L$, find b' so that $g(a)+h(b')=M$ and a' such that $g(a')+h(b)=N$. Then $(a',b')$ sits on the desired level set. Now one can easily engineer binary operations compatible with the usual order that don't have level sets that behave this way - indeed one may interpret the desired condition as a condition on the geometry of the level sets themselves!