Equivalent definition of abelian group Assuming that given operation is commutative and associative, are following definitions equivalent?
1) A zero element $e_0$ exists such that $e_i+e_0=e_i$ and inverse $e_{i'}$ exists for every $e_i$ such that $e_i+e_{i'}=e_0$
and
2) For every pair of elements $e_i$ and $e_j$ there exists an element $e_k$ such that $e_i+e_k=e_j$
 A: No, they're not equivalent. For example, let $S$ be any set, and let $+$ be the operation on $S$ given by $a + b = b$ for all $a,b\in S$. This satisfies (2) but not (1).
The issue here that you might be overlooking is that (1) says that there exists a single element $0$ which satisfies $a+0 = a$ for all $a$, while (2) only guarantees that for each $a$ there is some element $b$ such that $a + b = a$. It might be different for every $a$.
Now, you might complain that the $+$ operation in my example isn't commutative. If you mean to ask whether (1) and (2) are equivalent relative to some other axioms about $+$, you should specify exactly what axioms you have in mind.

Edit: Now that you've edited the question to add associative and commutativity of $+$, I can give a better answer. By the way, if $(S,+)$ is a set with an associative commutative operation, we call $S$ a commutative semigroup. Adding (1) makes $S$ an abelian group.
First, it's clear that (1) $\implies$ (2): Given $a$ and $c$ in an abelian group, and letting $b = (-a) + c$, we have $a + b = c$. In fact, we get something stronger, let's call it
(2'): For all $a$ and $c$ there exists a unique $b$ such that $a + b = c$. 
Indeed, if we have $a + b = c$, then adding $(-a)$ to both sides, we must have $b = (-a) + c$. 
Wikipedia calls (2') the Latin square property.
Now it turns out that also (2')$\implies$ (1). A binary operation satisfying (2') (for addition on the right and the left), but not necessarily associativity or commutativity, is called a "quasigroup", and every associative quasigroup is a group. The argument is simpler than the one in the link if we assume commutativity, so I'll give it here. 
Identity: Let $x$ be any element of $S$, and let $e$ be the unique element such that $x + e = x$. Now let $y$ be any other element. We have $x + (y+e) = (x + e) + y = x + y$, so both $b = y+e$ and $b = y$ satisfy the equation $x + b = x + y$. Again by uniqueness, $y + e = y$. So $e$ is an identity element for $(S,+)$.
Inverses: Now that we have $e$, (2') gives inverses easily. For any $x$, let $(-x)$ be the unique element of $S$ such that $x + (-x) = e$.
So we've established that (1) is equivalent to (2'). Is it equivalent to (2)? I don't actually know... But here's an observation: Given $(S,+)$ for all $a$, there is a function $f_a\colon S\to S$ given by $f_a(x) = a+x$, and if $S$ satisfies (2), then $f_a$ is surjective for all $a$. Since any surjective function from a finite set to itself is also injective, any finite commutative semigroup satisfying (2) must also satisfy (2'). So any counterexample to the equivalence of (1) and (2) must be infinite. Maybe someone else can provide one or a proof that (2) $\implies$ (2') for commutative semigroups.

Edit 2: A proof that (2) $\implies$ (2') has been given in an answer to my follow-up question.
A: I'll make an attempt to answer although algebra is not my strongest area..
First of all. A field has two operations. What you seem to be talking about are a requirements on one of these operations which obey the axioms of a group.


*

*Existance of zero or neutral element means there is one element
which if operated together with any other element yields that other element. This is a requirement for a group. However you clump it together with the inverse, which is a separate requirement. It is possible to have neutral elements without any inverse coupling between group elements.

*This is the closure requirement of a group. That the "result" of the operation between any two elements must stay in the group.
