Does $\exists e \in G \forall g \in G: e \circ g = g \circ e = g$ properly define the neutral element? I'd like to know if this
$$\exists e \in G \forall g \in G: e \circ g = g \circ e = g$$
is a real and accepted definition of a neutral element that is right and left identity. 
My textbook shows the following in the definition of a group:
$$\exists e \in G \forall g \in G: e \circ g = g$$ But later on there is the remark that the neutral element is "unique" and showing this: $$\forall g \in G: g \circ e = g$$
I'm still wrapping my head around mathematical notations so I hope this doesn't seem too trivial. Thanks in advance.
 A: Given a left identity, i.e. an element $e\in G$ such that $e\circ g=g$ for all $g\in G$, this one already is a right identity aswell. To see this, go for a contradiction. Suppose $g\circ e\ne g$, then
\begin{align*}
g\circ e&\ne g\\
(g\circ e)\circ g&\ne g\circ g\\
g\circ (e\circ g)&\ne g\circ g\\
g\circ g&\ne g\circ g\\
g&\ne g
\end{align*}
The last step follows from multplying with the (left) inverse $g^{-1}$. The contradiction implies that $g\circ e=g$ for every left identity $e$ (see ProofWiki for a simpler proof). Given a left identity is also a right identity, consider two such elements $e,e'\in G$. Then
$$e\overset{(1)}=e\circ e'\overset{(2)}=e'$$
In $(1)$ we use that $e'$ is a right identity and then in $(2)$ that $e$ is a left identity. Therefore the left identity (which happens to be also a right identity aswell) is uniquely defined. By the way, it is also sufficient to force every element to have a left inverse to obtain a group (there is a similiar proof showing that left inverses are also right inverses).
A: The typical definition is the one you provided. The definition stated in your textbook is that of a left identity, a concept which is weaker (in more general circumstances). Here's an example:

Take any non-empty set $A$, and define the binary operation
  $$a \circ b = b$$
  Then $\circ$ is associative, and every $a \in A$ is a left-identity (but not a right identity unless $|A| = 1$).

If $|A| > 1$, note that we fail to get uniqueness too.
I'm not entirely sure what the textbook is aiming for, but I can prove the following result:

Suppose $G$ has an associative binary operation $\circ$, and $e$ is a specific left identity. Further, suppose that for all $a \in G$, there exists some $b \in G$ such that $b \circ a = e$. Then $(G, \circ)$ is a group, and $e$ is its (two-sided) identity.

Proof. Suppose $a \in G$ and let $b \in G$ such that $b \circ a = e$. Then,
$$(a \circ b) \circ (a \circ b) = a \circ (b \circ a) \circ b = a \circ (e \circ b) = a \circ b.$$
Let $c \in G$ such that $c \circ (a \circ b) = e$, as guaranteed by the hypotheses. Then,
$$e = c \circ (a \circ b) = c \circ ((a \circ b) \circ (a \circ b)) = (c \circ (a \circ b)) \circ (a \circ b) = e \circ (a \circ b) = a \circ b.$$
So, if $e$ is a two-sided identity, then $b$ is a two-sided inverse for $a$. To prove $e$ is a two-sided identity, consider
$$a \circ e = a \circ (b \circ a) = (a \circ b) \circ a = e \circ a = a.$$
QED.
A: Your first definition is perfectly valid; it is simply that, as a result of the other axioms and itself, it implies the uniqueness of that element as well. Your textbook also leaves out the reversal of the elements, which is also implied. That said, since the properties left out are corollaries, they can be left out, so it is ultimately a matter of preference as to which you use. Personally I prefer the first.
For completeness, proofs that fill in the blanks for your text:

Uniqueness
Suppose $e$ is not unique; i.e. suppose $e'$ is another identity alongside $e$. Then
$$e = e \circ e' = e' \circ e = e'$$
The first equality is since $g \circ e = g$ for all $g$, and the second because $e \circ g = g$ for all $g$. But $e'$ is an identity, so the same must apply to it. After all, "for all $g$" include the identities.

Identity Commutes
Let $e$ be the identity, with $e \circ g = g$ for all $g$. We seek to show $g \circ e = g$ too. Note that
$$g = e \circ g = e \circ (g \circ e) = g \circ e$$
The first equality just is the given in the definition. The second follows because of the same: $g \circ e = g$. Of course, $(g \circ e) \in G$ so the axioms apply to it, and thus $e \circ (g \circ e) = g \circ e$.
A symmetrical argument works starting with $g \circ e = g$ instead (your text randomly flipped between the two).
A: Your definition is the right one, the other one is a property of groups that requires a proof. For instance, your definition applies as well to semigroups, but the second one would not be correct in this context. Consider for instance the semigroup $\{a,b\}$ defined by $ab=bb=b$ and $ba=aa=a$. Then both $a$ and $b$ are left identities but none of them is an identity.
