Is the set $\{0,1,2,3,4,5\}$, with the binary operation of "addition, then modulo $3$", a group? 
Consider a binary operation on the finite set $S=\{0,1,2,3,4,5\}$ where the operation is "addition, then modulo 3"  ($+_3$).
Does $S$ form a group under the binary operation $+_3$?

Now this binary operation follows closure property . It's associative. There exists identity element which is $0$. And for each element of the set $S$ there exists its inverse such that $aa^{-1}=0$.
It follows all requirements to be a group.
But it violates some other properties of group like left and right cancellation laws:
If $ax=ay\implies x=y$
$0+_32=0+_35\implies 2=5$
And equations $a+_3x=b$ and $y+_3a=b$ don't have unique solutions.

This is in fact, not a group. We can just check via brute force that it has no identity element:
$0$ is not the identity, since $3+_30=0\not=3$.
$1$ is not the identity, since $3+_31=1\not=3$.
$2$ is not the identity, since $3+_32=2\not=3$.
$3$ is not the identity, since $3+_33=0\not=3$.
$4$ is not the identity, since $3+_34=1\not=3$.
$5$ is not the identity, since $3+_35=2\not=3$.
Remember, for an element $e$ to be the identity, we would need that $x+_3e=x$ for every element $x$. But the above computations show that no such $e$ exists here.
