Edit: Sorry about misinterpreting. What my previous argument shows is that there is a counterexample if and only if there is a counterexample for $s=2$.

Unfortunately, there is a counterexample for $s=2$.

$(1-i-j-k)(1+i+j+k)=4=0$ mod $4$.

$(1+i+j+k)i(1-i-j-k)=(1-i-j-k)(i-1+k-j)=j^2 - (1-i-k)^2$

$= -1-1+1+1+2i+2k=2i+2k\neq 0$ mod $4$

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 $s$.

As in the regular quaternions, we have the formula $(a_0+a_1i+a_2j+a_3k)(a_0-a_1i-a_2j-a_3k)=a_0^2+a_1^2+a_2^2+a_3^2$. So since the quaternions are associative, as long as this is nonzero then $a_0+a_1i+a_2j+a_3k$ cannot be a zero-divisor, as $\mathbb Z_2$ has no zero divisors.

It is elementary to check that four squares cannot add to $0$. Without loss of generality at least one is odd. Every square is equal to $0,1$ or $4$ mod $8$, and there is no way to add a $1$ and three $0$s,$1$s, or $4$s to be $0$ mod $8$.

No zero divisors means semicommutative and reversible.

In other words, since the Hamiltonian quaternions are ramified at $2$, $\mathbb H \otimes \mathbb Q_2$ is not isomorphic to the matrix algebra over $\mathbb Q_2$, so by Artin-Wedderburn it's a division algebra.

Edit: Sorry about misinterpreting. What my previous argument shows is that there is a counterexample if and only if there is a counterexample for $s=2$.

Unfortunately, there is a counterexample for $s=2$.

$(1-i-j-k)(1+i+j+k)=4=0$ mod $4$.

$(1+i+j+k)i(1-i-j-k)=(1-i-j-k)(i-1+k-j)=j^2 - (1-i-k)^2$

$= -1-1+1+1+2i+2k=2i+2k\neq 0$ mod $4$