Question: Suppose I am given an equivalence relation on set A which is endowed with addition & closure under addition.

Then, to have addition on equivalence classes of A to be well-defined, it is to show:

$[a]+[b] = [a+b]$

Suppose I have shown that if $\forall a,b \in A, a \sim a', b\sim b'$ then $a+b \sim a'+b'$, then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$.

But then, now I need to show the other direction. I found it is easiest to define $c \in A$ so $b+c = 0$. So if $x \in [a+b]$ then $x \sim a+b$, and using what I proved, $x+c \sim a+(b+c) = a$. Letting $r = x-b$ and $s = b$ shows the other direction.

My question now is, to do this, (A,+) seems to need to have the property that all elements must have an identity, an inverse, and is associative, so what exactly must (A,+) be for all this to work, a group? abelian group? and why?

Note: Also I think I'm failing to consider a lot of things when thinking about this, so it will be very helpful if anyone could point many of those things out.

Answer: 
$(A,+)$ does not *need* to be a group. For example, we may take $A=\Bbb N$ (with usual addition) and define $a\sim b$ as "$a$ and $b$ end in the same decimal digit". Then $(A/{\sim},+)$ is a group (cyclic of order $10$).

Similary, $(A/{\sim},+)$ may be an abelian group even if $(A,+)$ is non-abelian. (But not vice versa).
If $A$ is a set, $+$ is a binary operation on $A$ and $\sim$ denotes an equivalence relation on $A$ then we can define a binary operation on $A/\sim$ by means of:$$[a]+[b]:=[a+b]\tag1$$if and only if $$\forall a,a',b,b'\in A [a\sim a'\wedge b\sim b'\implies a+b=a'+b']\tag2$$

> 
> "...then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$..."
> 
> 
> 

No, $(1)$ tells us directly that $[a+b]=[a]+[b]$ and there is no such thing needed as "..showing the other direction..".

The only thing you really need is a proof that $(2)$ is valid. It is actually the proof that the operation "defined" in $(1)$ is indeed well defined.