Assume that exists a zero element, say $e$. Then for every $a\in\mathbb{Z}$ must hold $a\ast e=a=e\ast a$. So, if we choose an element $b\neq e$ the same must hold, i.e. $b\ast e=b=e\ast b$. But then, by definition, must hold $b=b=e$. Contradiction. So, the operation $\ast$ does not define a group's structure on $\mathbb{Z}$.
P.S. By this specific operation, in any set with more than one elements we can not have a group's structure. It can be proved using the same argument.
_________________ Grigorios Kostakos
