Groups, if $ab = cd$ then $axb = cxd$ for all groups? Let $G$ be a group such that $ a,b,c,d,x ∈ G $ with $ a ≠ c , b ≠ d $
if $ ab = cd $ then $ axb = cxd $ for all groups?
 A: Let $b = e$, the identity element. Then we have $a = cd$, which implies 
$$ad^{-1}c^{-1} = 1$$
Next, if we have
$$axb = cxd \Longrightarrow ax = cxd$$
We can take $x = a^{-1}$, so that
$$1 = ca^{-1}d$$
So that
$$ ad^{-1}a^{-1}d = (ad^{-1}c^{-1})(ca^{-1}d) = (1)(1) = 1$$
But if $ad^{-1}a^{-1}d = 1$, then $ad^{-1} = d^{-1}a$, so $a$ and $d^{-1}$ commute, so $a$ and $d$ commute. 
So the question is: must $a$ and $d$ commute if $a = cd$? Surely not, for if $d$ and $a$ do not commute, we take $c = ad^{-1}$ and the above equation works.
A: The conditions given fail to rule out a particularly trivial approach, which in fact works for monoids (semigroups with identity) as well as groups:
Suppose that for all $a,b,c,d,x\in G$ such that $a\ne c$, $b\ne d$, and $ab=cd$, $axb = cxd$.
Let $a=d=p$ and $b=c=e$, where $p$ is an arbitrary non-identity element and $e$ is the identity. Note that indeed $a\ne c$, $b\ne d$, and $ab=cd=p$.
Then for arbitrary $x$, $axb=cxd$, so $pxe=exp$, and thus $px=xp$. If $p=e$, then of course this holds as well. So $G$ is abelian.
A: If $b=c=1$, then $a=d$, so your claim implies that $ax=xd=xa$, which fails in nonabelian groups.
