Assume $xy=0$ but $yx\neq 0$. We can take $x$ and $y$ to be $2$-adic quaternions, and then we have $xy=0$ modulo $2^s$ but $yx\neq 0$ modulo $2^s$. We can factor out the highest possible power of $2$ from $x$ and $y$, and then divide $2^s$ by that power, so without loss of generality $x$ and $y$ are nontrivial modulo $2$.
$xy\bar{y}=x|y|^2$. $|y|^2$ is a regular $2$-adic number and is the sum of four squares, at least one odd, and so is nonzero mod $8$. $x$ has a coefficient that is nonzero mod $2$ so that same coefficient in $x|y|^2$ is nonzero mod $8$, so $x|y|^2$ is nonzero mod $8$, so $xy$ is nonzero mod $8$.
Therefore $s$ must be $1$ or $2$. Since a-fortiori computed that reversibility holds for $s=1,2$, it holds for every $s$.