Proving the validity of a group structure on $\mathbb{Z} \times \mathbb{Z}$.
The identity element need not be $(0,0)$. It should be $(x,y)$ such that, for every $(a,b)$, $(a,b)\cdot(x,y)=(a,b)$ and $(x,y)\cdot(a,b)=(a,b)$.
Take $a=b=1$ and you immediately get a contradiction: $$ (1,1)\cdot(x,y)=(2,x+y)\ne(1,1) $$ So the operation lacks an identity and you're finished. No need to check associativity.