Let, $ab=e\land bc=e\tag {1}$
for some $b,c\in G$. And, $ae=a\tag{2}$
From $(2)$, $$eae=ea\implies(ab)a(bc)=ea\implies ((ab)(ab))c=ea\implies ec=ea\tag{3}$$
Similarly, $$ae=a\implies a(bc)=a\implies (ab)c=a\implies ec=a\tag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=e\implies b((ab)(bc)a)=e\implies ba=e$$
The statement of the associativity condition is wrong; it should be
$$(a \ast b) \ast c = a \ast (b \ast c) .$$
Expanding the l.h.s. gives
\begin{align}(a \ast b) \ast c &= (a + b - ab) \ast c \\ &= (a + b - ab) + c - (a + b - ab) c \\ &= a + b + c - bc - ca - ab + abc .\end{align}
Now, do the same thing for the r.h.s. and verify that the expressions agree.
We cannot simply assume the existence of an identity and inverse. There are plenty of associative binary operations which lack one or both!
We can pick out the identity element using the definition: We must have, for example, that
$$a \ast e = a$$ for all $a$, and expanding gives
$$a + e - ae = a.$$
Can you find which $e$ makes this true for all $a$?
Likewise, to show that the operation admits inverses, it's enough to produce a formula for the inverse $a^{-1}$ of an arbitrary element $a$, that is, an element that satisfies $a \ast a^{-1} = e = a^{-1} \ast a$. As with the identity, use the definition of $\ast$ and solve for $a^{-1}$ in terms of $a$.
Remark If we glance at the triple product, which by dint of associativity we may as well write $a \ast b \ast c$, we can see the occurrence of the elementary symmetric polynomials in $a, b, c$, which motivates writing that product as $(a - 1) (b - 1) (c - 1) + 1$. Then, glancing back we can see that we can similarly write $a \ast b = (a - 1) (b - 1) + 1,$ which is just the conjugation of the usual multiplication (on $\Bbb R - \{0\}$) by the shift $s : x \mapsto x + 1$, that is, $a \ast b := s(s^{-1}(a) s^{-1}(b))$. Since multiplication defines a group structure on $\Bbb R - \{0\}$, that $\ast$ defines a group structure follows by unwinding definitions. For example, to show associativity, we have
$$(a \ast b) \ast c = s(s^{-1}(s(s^{-1}(a) s^{-1}(b))) s^{-1}(c)) = s(s^{-1}(a) s^{-1}(b) s^{-1}(c)) = s(s^{-1}(a)s^{-1}(s(s^{-1}(b) s^{-1}(c))) = a \ast (b \ast c) .$$
(Note that in writing the third expression without parentheses we have implicitly used the associativity of the usual multiplication.) You can just as well use this characterization to determine the group identity $e$ and the formula for the inverse of an element under $\ast$.
Indeed, this together with analogous arguments for the other group axioms shows that conjugating any group operation by a set bijection defines an operation on the other set.
Best Answer
Consider the example of $\ast$ such that $x\ast y = x$ for all $x,y$ over some set $G$ with at least two distinct elements.
Clearly, this operation is closed.
Clearly, there exists a right-identity (and in fact, everything is a right identity) $e$ such that $x\ast e = x$ for all $x$.
It should also be clear that this operation is infact associative as $(x\ast y) \ast z = x\ast z = x = x\ast y = x\ast (y\ast z)$
However, so long as $G$ has at least two distinct elements should be clear that there is no left-identity as for any proposed left identity $\ell$ and $x$ not equal to $\ell$, you would have $\ell \ast x = \ell$, not $x$.
This all shows that the conditions of closure, associativity, and existence of right identity alone are not enough to prove the existence of left-identity. For that, one would need additional assumptions.