Do these conditions force $G$ to be a group? Let $G$ be a set $G=\{e,g,g'\}$ such that $e\circ{} g = g$ and $g\circ g'=e$ where $\circ$ is an associative operation. These conditions are seemingly weaker than outright stating $G$ is a group because both the identity and the inverses only work from one side. However, we should be able to show that these conditions force $G$ to be a group. 
That is, using these conditions, we should be able to show that $g\circ e=g$ and $g'\circ g =e$. 
I have gone round and round trying to get this to work with no avail. I'm pretty sure we can't use the cancellation property for groups since we haven't shown that $G$ is a group yet. 
Any ideas?
 A: If no further conditions are given, then you cannot prove that $G$ is a group because you've only given 2 entries in the multiplication table and you need 9.
If you assume that $\circ$ is associative, then you can deduce more. For instance:
$$
e \circ e = e\circ (g \circ g') = (e\circ g) \circ g' = g \circ g' = e
$$
which gives you 1 more entry in the table.
Associativity restricts the multiplication table but there's still lots of room. See for instance


*

*Associative Operations on a Three-Element Set

*How many binary operations are associative?
A: It is not enough (see comment of André and answer of lhf). In this context it is interesting to note that the following conditions are indeed enough for $G$ to be a group:
1) associative multiplication
2) existence of left-identity $e$ (i.e $eg=g$ for each $g$)
3) existence of left-inverse $g^{-1}$ for each $g$ (i.e. $g^{-1}g=e$ for each $g$)
If $x^2=x$ then it can be shown under these conditions that $x=e$. This by: $$x=ex=x^{-1}x^2=x^{-1}x=e$$ So from $gg^{-1}gg^{-1}=geg^{-1}=gg^{-1}$ it follows that $gg^{-1}=e$. This shows that $g^{-1}$ also works as a right-inverse. Then you have $ge=gg^{-1}g=eg=g$ showing that $e$ also works as right-identity.
If the word 'left' in 2) and 3) is replaced by 'right' then you also have sufficient conditions.
A: I had made some huge mistakes here. The problem stated that for every $g\in G$ there exists some $g'\in G$ such that $gg'=e$. I mistakenly took this to mean that those were the only elements in $G$ (hey, it's late). But if I let $gg'=e$ and $g'g''=e$ then I can say
$g=g*e=g*(g'*g'')=(g*g')*g''=g'*g''=e$
$g'*g=g'*(e*g'')=(g'*e)*g''=g'*g''=e$
So we now have $g*g'=g'*g=e$ for all $g\in G$
Then
$e*g=(g*g')*g=g*(g'*g)=g*e=g$
Giving (finally!), 
$g*e=e*g=g$. 
Now it's a group.
