A binary operation on $G$ satisfies $a\cdot(b\cdot c)=(a\cdot b)\cdot (a\cdot c)$ and has an identity, is it a group?

Let $e$ denote the identity element. Note that $$xy=x(ey)=(xe)(xy)=x(xy)$$ and $$xy=x(ye)=(xy)(xe)=(xy)x$$ for any $x,y\in G$. Now we have $$a(bc)=(ab)(ac)=((ab)a)((ab)c)=(ab)((ab)c)=(ab)c,$$ proving associativity (the first two equalities are the given axiom, and the second two come from $(xy)x=xy$ and $x(xy)=xy$ for appropriate choices of $x$ and $y$).

However, $G$ does not have to be a group. For instance, $G$ could be any totally ordered set with a greatest element, with the operation given by $ab=\min(a,b)$; this is not a group unless $G$ has only one element, since there are no inverses. Or more generally, $G$ could be any partially ordered set in which any finite subset has a greatest lower bound, with $ab$ being the greatest lower bound of $a$ and $b$. In fact, every commutative example is of this form (explicitly, you can define $a\leq b$ iff $ab=a$ and then $ab$ will be the greatest lower bound of $a$ and $b$ with respect to this order).

For an example that is not commutative, consider $G=\{e,a,b\}$ with the operation given by $xy=x$ unless $x=e$ in which case $xy=y$.

If it is a gruop, then it must be trivial: Infact if you fix $a\in G$ then for each $b,c$ you have that

$a(bc)=(ab)(ac)$

but the operation is associative, so

$(ab)c=a(bc)=(ab)(ac)$

For $b=1,c=1$ you have

$a=a^2$

But $G$ is a group, so

$a=a^2a^{-1}=aa^{-1}=1$

Thus

$G=\{1\}$

@JMoravitz already showed we can't get a non-trivial group this way, but also that $\cdot$ is idempotent. Associative non-group options include taking $\cdot$ to be $\land$ or $\lor$ on propositions, or similarly $\cap$ or $\cup$ on sets. The respective identities are True, False, the universe or $\emptyset$.