One surprisingly nice axiomatization for this operator is:
$$x/x=1$$
$$x/1=x$$
$$(x/z)\,/\,(y/z)=x/y$$

The proof is straightforward, and most easily dealt with by writing it out in detail:

We translate sentences from group theory with $(x^{-1})^*=1/x$, $(xy)^*=x/(1/y)$.

The first two axioms of group theory then are

\begin{align}
a1=a&: a\,/\,(1/1)=a/1=a\\
1a=a&: 1\,/\,(1/a)=(a/a)\,/\,(1/a)=a/1=a\\
aa^{-1}=1&: a\,/\,(1/(1/a))=a/a=1\\
a^{-1}a=1&: (1/a)\,/\,(1/a)=1.\\
\end{align}

Associativity is
$$(ab)c=a(bc): (a/(1/b))\,/\,(1/c)=a\,/\,(1/(b/(1/c)))$$
which is a special case of
\begin{align}
(a/(1/b))\,/\,d
&=(a/(1/b))\,/\,(d/1)\\
&=(a/(1/b))\,/\,((d/b)/(1/b))\\
&=a\,/\,(d/b)\\
&=a\,/\,((d/d)/(b/d))\\
&=a\,/\,(1/(b/d)).
\end{align}