'Inverse' property of a group and the special case that makes a group an Abelian group One of the property for which a set must have in order to be a group is to possesses the 'inverse' property.
What this says is that for each element $a$ in $G$, there is an element $b$ in $G$ with $ab = ba = e$ where $e$ is the identity element.
In the special case where, in addition to the above 'inverse' property, if $ab = ba$ for all $a,b \in G$, we defined $G$ to be an Abelian group.
My confusion lies in the difference between the logical interpretation of the 2 properties: 
Am I correct to interpret " for each element $a$ in $G$" as being equivalent to "for all element $a$ in $G$"?
In other words, 
Let $G$ be the set $G=\{a_1,a_2,...,b,...,a_n\}$ and $b$ is unique
Then, by the inverse property, we have $\{a_1,a_2,...,a_n\}.b=e$.
secondly, what is the difference between " there is an element $b$ in $G$" and "for all element $b$ in $G$"?
 A: One has to be careful about the ORDER (sequence) of logical quantifiers. The usual way the existence of inverses in a group is quantified is like so:
$\forall a \in G,\ \exists b \in G: a\ast b = b\ast a = e_G$.
This means we get a $b$ that may very well (and it turns out indeed does) depend on $a$.
Your "interpretation" is the following:
$\exists b \in G: \forall a \in G, a\ast b = b\ast a = e_G$
Notice how "swapping" the order of the quantifiers now says something completely different. Now it says that a SINGLE $b$ will work for ANY (and every) $a$.
Of course, the "second version" is the language (or logical version thereof) used to describe the identity:
$\exists e \in G: \forall a \in G, a\ast e = e\ast a = a$.
So your confusion is understandable, both the identity axiom and the inverse axiom of group theory are both labelled: "existence of...", and the niceties of the distinction of the two KINDS of existence are often glossed over.
Put another way: the identity element is a distinguished element of $G$, that acts on all other elements of $G$ in the same way: leaving them unchanged via multiplication. Inverses, on the other hand, only "undo" (or invert) what they are inverses OF, their scope is limited.
It is possible, in some algebraic structures you need not worry about just yet, for identities and/or inverses to be "only one-sided", and this can make it possible for them to be non-unique. However, in groups, this cannot happen: any group identity, is necessarily unique, for if we had another, say $e'$, then:
$e = e\ast e' = e'$.
In a similar vein, if an element $a$ had two, two-sided inverses, say $b,b'$, then:
$b = e\ast b = (b' \ast a)\ast b = b'\ast (a \ast b) = b' \ast e = b'$
We can say even more: suppose $a \ast b = a' \ast b = e$ (that is, $b$ is a right-inverse to $a$ and $a'$). If we are in a group, we know that $b$ has an inverse $b'$, so:
$(a \ast b) \ast b' = (a' \ast b)\ast b'$
$a\ast(b\ast b') = a'\ast(b\ast b')$
$a\ast e = a'\ast e$
$a = a'$, so $b$ can only be an inverse to a SINGLE group element.
In short, we have shown that for ANY group $G$, the mapping:
$g \mapsto g^{-1}$ is a BIJECTION (one-to-one and onto).
A: To answer your first question, the important thing is to realize that different element may have different inverse. For example, $(\mathbb{Z},+)$ is a group. Consider $3$ and $5$ in $\mathbb{Z}$. It is correct to say that these two elements have inverse. But keep in mind that their inverses are different! Indeed, $3$ has inverse $(-3)$, and $5$ has inverse$(-5)$.
To answer your second question, no, they are not the same.
A: The inverse property says that for each $a\in G$ you can find at least one element $b \in G$ such that $ab=ba=e$. You can now use the associativity to show that it must be unique, since if $ac=ca=e$ also, then $c=ce=c(ab)=(ca)b=eb=b$. But if I choose a different $a'\in G$ , then I will get a different inverse element $b'\in G$ such that $a'b'=b'a'=e$, and it is not true that $a'b=e$ (and in particular that $\{a_1 ,...,a_n\}\cdot b \neq \{e\}$. 
Think for example inside the group of integers. For every $n\in \mathbb{Z}$ there is exactly one integer $m$ such that $m+n=n+m=0$, namely $m=-n$. When I go to $n+1$ its inverse is $-(n+1) \neq -n$. 
When you say something like "for each element $a\in G$" some "statement" $P(a)$ happens, it means that if I choose some $a\in G$ no matter which one, the statement $P(a)$ is true. For example "P(a)= there is at least one element $b$ such that $ab=e$". Notice that the $b$ is chosen according to $a$. Another example - "For each rational number $q$, we can find another rational $r$ such that $2r=q$.
