Let $(L, <)$ be any linear order. Fix an arbitrary well-order $\sqsubset$ on $L$. For any $a\le b$ define $m(a,b)=m(b,a):= $ the $\sqsubset$-least element in the interval $[a,b]$.

Then $m$ is associative and commutative. (But rather uninteresting from the point of view of the relation $ < $.)

Notation: Write $\langle x,y\rangle:=\{ z: x\le z\le y \text{ or } y\le z \le x\}$.

Proof of associativity: Let $a,b,c$ be arbitrary, and let $d$ be the $\sqsubset$-least element in the interval $J:=[\min(a,b,c),\max(a,b,c)]$. I claim that $m(m(a,b),c)=d$. As $m$ is commutative, and $\min$ and $\max$ are commutative and associative, this also implies $m(a,m(b,c))=d$.

• Case 1: $d\in \langle a,b\rangle$. Then $m(a,b)=d$ (as $\langle a,b\rangle \subseteq J$), and $m(m(a,b),c)=m(d,c)=d$, as also $\langle d,c\rangle \subseteq J$.
• Case 2: $d\notin \langle a,b\rangle$. Wlog $a\le b$. We must have $c\notin [a,b]$, wlog $b< c$. So $d\in [b,c]$.
  Let $d':=m(a,b)$, then $d\sqsubset d'$, and $m(m(a,b),c) = m(d',c) = d$, as $d\in [b,c]\subseteq [d',c] = \langle d',c\rangle \subseteq J$.

QED.

Let $(L, <)$ be any linear order. Fix an arbitrary well-order $\sqsubset$ on $L$. For any $a\le b$ define $m(a,b)=m(b,a):= $ the $\sqsubset$-least element in the interval $[a,b]$.

Then $m$ is associative and commutative. (But rather uninteresting from the point of view of the relation $ < $.)

Notation: Write $\langle x,y\rangle:=\{ z: x\le z\le y \text{ or } y\le z \le x\}$.

Proof of associativity: Let $a,b,c$ be arbitrary, and let $d$ be the $\sqsubset$-least element in the interval $J:=[\min(a,b,c),\max(a,b,c)]$. I claim that $m(m(a,b),c)=d$. As $m$ is commutative, and $\min$ and $\max$ are commutative and associative, this also implies $m(a,m(b,c))=d$.

• Case 1: $d\in \langle a,b\rangle$. Then $m(a,b)=d$ (as $\langle a,b\rangle \subseteq J$), and $m(m(a,b),c)=m(d,c)=d$, as also $\langle d,c\rangle \subseteq J$.
• Case 2: $d\notin \langle a,b\rangle$. Wlog $a\le b$. We must have $c\notin [a,b]$, wlog $b< c$. So $d\in [b,c]$.
  Let $d':=m(a,b)$, then $d\sqsubset d'$, and $m(m(a,b),c) = m(d',c) = d$, as $d\in [b,c]\subseteq [d',c] = \langle d',c\rangle \subseteq J$.

QED.

(Edit:) It is also easy to see that $m$ is monotone. But in general not continuous.

Let $(L, <)$ be any linear order. Fix an arbitrary well-order $\sqsubset$ on $L$. For any $a\le b$ define $m(a,b)=m(b,a):= $ the $\sqsubset$-least element in the interval $[a,b]$.

Then $m$ is associative and commutative. (But rather uninteresting from the point of view of the relation $ < $.)

Notation: Write $\langle x,y\rangle:=\{ z: x\le z\le y \text{ or } y\le z \le x\}$.

Proof of associativity: Let $a,b,c$ be arbitrary, and let $d$ be the $\sqsubset$-least element in the interval $J:=[\min(a,b,c),\max(a,b,c)]$. I claim that $m(m(a,b),c)=d$, and also (by a dual proof) $m(a,m(b,c))=d$.

• Case 1: $d\in \langle a,b\rangle$. Then $m(a,b)=d$ (as $\langle a,b\rangle \subseteq J$), and $m(m(a,b),c)=m(d,c)=d$, as also $\langle d,c\rangle \subseteq J$.
• Case 2: $d\notin \langle a,b\rangle$. Wlog $a\le b$. We must have $c\notin [a,b]$, wlog $b< c$. So $d\in [b,c]$.
  Let $d':=m(a,b)$, then $d\sqsubset d'$, and $m(m(a,b),c) = m(d',c) = d$, as $d\in [b,c]\subseteq [d',c] = \langle d',c\rangle \subseteq J$.

QED.

Let $(L, <)$ be any linear order. Fix an arbitrary well-order $\sqsubset$ on $L$. For any $a\le b$ define $m(a,b)=m(b,a):= $ the $\sqsubset$-least element in the interval $[a,b]$.

Then $m$ is associative and commutative. (But rather uninteresting from the point of view of the relation $ < $.)

Notation: Write $\langle x,y\rangle:=\{ z: x\le z\le y \text{ or } y\le z \le x\}$.

Proof of associativity: Let $a,b,c$ be arbitrary, and let $d$ be the $\sqsubset$-least element in the interval $J:=[\min(a,b,c),\max(a,b,c)]$. I claim that $m(m(a,b),c)=d$. As $m$ is commutative, and $\min$ and $\max$ are commutative and associative, this also implies $m(a,m(b,c))=d$.

• Case 1: $d\in \langle a,b\rangle$. Then $m(a,b)=d$ (as $\langle a,b\rangle \subseteq J$), and $m(m(a,b),c)=m(d,c)=d$, as also $\langle d,c\rangle \subseteq J$.
• Case 2: $d\notin \langle a,b\rangle$. Wlog $a\le b$. We must have $c\notin [a,b]$, wlog $b< c$. So $d\in [b,c]$.
  Let $d':=m(a,b)$, then $d\sqsubset d'$, and $m(m(a,b),c) = m(d',c) = d$, as $d\in [b,c]\subseteq [d',c] = \langle d',c\rangle \subseteq J$.

QED.