Let $M$ be a commutative monoid with the cancelation law and suppose that $a \nsim b, x \nsim y, ax = by, ay = bx$, and $a$ and $b$ are irreducible. A first question was to show that $\gcd(ax,bx) = \emptyset$. Then, we have to show that if $a \nmid y$ or $b \nmid x$, then the set of least common multiples of $(a,b)$ is the empty set. I managed to do the first one after a lot of thinking, but I can't understand the second one. I tried a proof by contradiction, but I can't get to a contradiction.
The proof for the first part goes like this:
Suppose that $\gcd(ax,bx) \neq \emptyset$. Then, $\gcd(ax,bx) = x \gcd(a,b) = x U(M)$, where $U(M)$ are the units of the monoid. Then, $a = xu$ and $b = xu^\prime$, thus $ax=by$ leads to $x \sim y$, which is a contradiction.
EDIT: Originally I had said this was an integral domain, it is actually a monoid, only the operation of multiplication exists.
Note that for any $z\in M$, if $x,y$ are replaced by $xz,yz$, the hypotheses of the problem are still satisfied.
Thus your assertions that $a\sim x$ and $b\sim x$ are not justified.
Instead, we can argue as follows . . .
Assume that $\gcd(ax,bx)$ exists.
Our goal is to derive a contradiction.
From the equations $$ \left\lbrace \begin{align*} ax&=by\\[4pt] ay&=bx\\[4pt] \end{align*} \right. $$ we get $abx^2=aby^2$, so $x^2=y^2$.
As you noted, since $\gcd(a,b)=1$, we get $$ \gcd(ax,bx)=x{\,\cdot\,}\gcd(a,b)=x $$ Then from $ax=by$ we get $y{\,\mid\,}ax$, and from $ay=bx$ we get $y{\,\mid\,}bx$, so $y$ is a common divisor of $ax$ and $bx$.
Hence $y{\,\mid}\gcd(ax,bx)$, so $y{\,\mid\,}x$.
Then $x=yz$ for some $z\in M$, hence \begin{align*} & x^2=y^2 \\[4pt] \implies\;& x(yz)=y^2 \\[4pt] \implies\;& xz=y \\[4pt] \implies\;& x{\,\mid\,}y \\[4pt] \end{align*} But then we have $x{\,\mid\,}y$ and $y{\,\mid\,}x$, so $x\sim y$, contradiction.
Therefore $\gcd(ax,bx)$ does not exist.
Next we show that $\text{lcm}(ax,bx)$ does not exist.
The following lemma will clinch it . . .
Lemma:$\;$If $s,t\in M$ and $\text{lcm}(s,t)$ exists, then $\gcd(s,t)$ exists.
Proof of the lemma:
Let $m=\text{lcm}(s,t)$.
Let $g,h\in M$ be such that $m=gs$ and $m=ht$.
Since $st$ is a common multiple of $s$ and $t$, it follows that $st=em$ for some $e\in M$.
Then $$em=st\implies e(ht)=st\implies eh=s\implies e{\,\mid\,}s$$ and $$em=st\implies e(gs)=st\implies eg=t\implies e{\,\mid\,}t$$ so $e$ is a common divisor of $s$ and $t$.
Now let $d$ be any common divisor of $s$ and $t$.
Then $s=ds_1$ and $t=dt_1$ for some $s_1,t_1\in M$.
Now $ds_1t_1$ is a common multiple of $s$ and $t$, hence $ds_1t_1=km$ for some $k\in M$, so then \begin{align*} & st=em \\[4pt] \implies\;& (ds_1)(dt_1)=em \\[4pt] \implies\;& (d)(ds_1t_1)=em \\[4pt] \implies\;& (d)(km)=em \\[4pt] \implies\;& dk=e \\[4pt] \implies\;& d{\,\mid\,}e \\[4pt] \end{align*} hence $\gcd(s,t)=e$, which completes the proof of the lemma.
Since we've already shown that $\gcd(ax,bx)$ does not exist, it follows from the lemma that $\text{lcm}(ax,bx)$ does not exist, as was to be shown.