The title is the main question.
Explaining the notation, an ordered group is a group $X$ with a partial order such that for every $x,y,z\in X$, if $x\leq y$, then $zx\leq zy$ and $xz\leq yz$.
An element $x\in X$ is called irreducible iff $e<x$ ($e$ is the neutral element) and for every $y\in X$, if $e\leq y\leq x$, then $y=e$ or $y=x$.
I already know that for any irreducible elements $x,y\in X$, if they have infimum $x\sqcap y$, then they commute.
Indeed, we have $e\leq x\sqcap y\leq x$, so $x\sqcap y=e$ or $x\sqcap y=x$. If $x\sqcap y=x$, then $e<x\leq y$, so $x=y$, so they commute. If $x\sqcap y=e$, then for every $z\in X$ we have
$(xy\leq z\Leftrightarrow e\leq x^{-1}zy^{-1}\Leftrightarrow yz^{-1}x\leq e\Leftrightarrow (yz^{-1}x\leq x\text{ and }yz^{-1}x\leq y)\Leftrightarrow (yz^{-1}\leq e\text{ and }z^{-1}x\leq e)\Leftrightarrow (y\leq z\text{ and }x\leq z))$
and analogously we have
$(yx\leq z\Leftrightarrow(x\leq z\text{ and }y\leq z))$
so both $xy$ and $yx$ are suprema of $x$ and $y$, so they are equal.
However, I have no idea about what to do if they do not have infimum.
The answer is No.
I will explain why, if $\Gamma$ is any nonabelian group that has a surjective group homomorphism $\varphi: \Gamma\to \mathbb Z$, there is a partial order on $\Gamma$ compatible with left/right multiplication for which some irreducible elements fail to commute.
Assume that $\Gamma$ is any nonabelian group, and that $\varphi: \Gamma\to \mathbb Z$ is a surjective group homomorphism. Define a strict partial order on $\Gamma$ by:
$F<G$ in $\Gamma$ iff $\varphi(F)<\varphi(G)$ in $\mathbb Z$
Then, of course, $F\leq G$ is defined to mean $F<G$ or $F=G$. This is a partial order on $\Gamma$, and it is compatible with left and right multiplication by elements of $\Gamma$, because the strict order on $\mathbb Z$ is compatible with left/right addition by elements of $\mathbb Z$.
The elements in $\varphi^{-1}(1)$ are irreducible in the order on $\Gamma$. The set $\varphi^{-1}(1)$ generates $\Gamma$, since $1$ generates $\mathbb Z$, so if $\Gamma$ is nonabelian then some pair of elements in $\varphi^{-1}(1)$ must fail to commute. This explains why $\Gamma$ must have a pair of noncommuting irreducible elements.
For example, if $\Gamma = F_{\textrm{Grp}}(x,y)$ is the free group over $\{x,y\}$, then we can take $\varphi:\Gamma\to \mathbb Z$ to be the map determined by $x\mapsto 1, y\mapsto 1$. Then both $x$ and $y$ are irreducible in the derived partial order, but they do not commute.
For a solvable example, take $\Gamma$ to be the group of functions $F\colon \mathbb R\to \mathbb R$ which are expressible as $F(x) = ax+b$ where $a$ is an integral power of $2$ and $b$ is a dyadic rational, and the multiplication operation of $\Gamma$ is composition of functions. Take $\varphi: \Gamma\to \mathbb Z$ to be the homomorphism $ax+b\mapsto \log_2(a)$. The functions $f(x)=2x$ and $G(x)=2x+1$ are irreducible in the derived ordering, but do not commute under composition.