I am thankful of Anton Klyachko who introduced axiomatic rank to me: the axiomatic rank of a variety is the minimum number of variables which we need to define that variety by identities. It seems clear that the axiomatic rank of the variety of groups is equal to three. But why?

Let $V$ be the variety axiomatized by \begin{align*} x+y&=y+x,\\ x+(x+y)&=y,\\ x+0&=x,\\ x&=x. \end{align*} (This also implies $x+x=0$.) Straightforward induction on the length of a term shows that every $2$variable term is equal over $V$ to one of $0,x,y,x+y$, which implies that $V$ axiomatizes all $2$variable identities valid in abelian groups of exponent $2$. If we can find a nonassociative algebra $A\in V$, it will follow that no variety of associative algebras (in the signature of groups) containing the group of order $2$ (which generates the variety of abelian groups of exponent $2$) can be axiomatized by identities in less than $3$ variables. I can’t think of a clever way to construct $A$ at the moment, but the following (which is just the $3$variable term algebra modulo the theory of $V$ in slight disguise) works. Let $A$ consist of a special element $0$ together with all finite unordered binary trees whose leaves are labelled $x$, $y$, or $z$, such that the two children of any nonleaf node are distinct (nonisomorphic), and if $a$ has children $b,c$, and $c$ is not a leaf, then its children are distinct from $b$. Define an operation $+$ on $A$ by $t+0=0+t=t$; $t+t=0$; if $s$ is a child of $t$, then $t+s=s+t$ is the other child of $t$; in the remaining cases, $t+s$ is the tree with a root whose children are $t$ and $s$. Then one can check that $A\in V$, and it is clearly nonassociative. 

