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!

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 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.

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!

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!