Prove an alternative definition of group (which replaces the identity & inverse axioms with another) I learned (from multiple sources such as http://planetmath.org/alternativedefinitionofgroup) that an alternative definition of group is obtained by replacing the identity and inverse axioms with the following axiom (please assume that all the quantifiers below are restricted to G):
$$\forall a,b \, \exists x,y :\, xa = ay = b.$$
The PlanetMath article mentioned above offers a proof (that this alternative does define group).  But a part of the proof, proving the existence of an identity, is cryptic, and, if I'm not mistaken, seems to have a whole.  Namely, that part seems to depend on the following:
$$\text{(H)} \quad If \quad \forall a,b \, \exists e:\, ea = ae = a \land eb = be = b, \quad then,\quad \exists e \, \forall a: \, ea = ae = a.$$
I haven't been able to show that this (H) follows from the alternative definition.  So, I have two questions, with the second depending on the affirmative answer to the first.  
 1. Is my reading (shown below) of that part of the PlanetMath proof correct?
 2. How does (H) follow from the alternative definition?

Any help will be much appreciated. Thank you.


My reading of the proof:

 1. Let $a,b \in G$ be any.
 2. There is $e_a \in G$ s.t. $\quad e_aa = a$.
$\quad$ (Because given $a,a \in G$, there is(are) $e_a, e_{a'} \in G$ s.t. $\quad e_aa = ae_{a'} = a$.)
 3. There is $e_b \in G$ s.t. $\quad be_b = b$.
$\quad$ (Because given $b,b \in G$, there is(are) $e_{b'}, e_b \in G$ s.t. $\quad e_{b'}b = be_b = b$.)
 4. $e_a = e_b$ because:
$\quad$ 4.1. There is $x \in G$ s.t. $\quad xb = e_a$.
$\quad$ (Because given $b,e_a \in G$, there is(are) $x,x' \in G$ s.t. $\quad xb = bx' = e_a$.)
$\quad$ 4.2. There is $y \in G$ s.t. $\quad ay = e_b$.
$\quad$ (Because given $a,e_b \in G$, there is(are) $y',y \in G$ s.t. $\quad y'a = ay = e_b$.)
$\quad$ 4.3. $e_a = xb \quad$ (by 4.1)
$\quad$ 4.4. $\quad = x(be_b) \quad$ (by 3)
$\quad$ 4.5. $\quad = (xb)e_b \quad$ (by associativity)
$\quad$ 4.6. $\quad = e_ae_b \quad$ (by 4.1)
$\quad$ 4.7. $\quad = e_a(ay) \quad$ (by 4.2)
$\quad$ 4.8. $\quad = (e_aa)y \quad$ (by associativity)
$\quad$ 4.9. $\quad = ay \quad$ (by 2)
$\quad$ 4.10. $\quad = e_b \quad$ (by 4.2)
 5. $\forall a,b \, \exists e:\, ea = a \land be = b. \quad$ (by 1-4)
 6. Let $a,b \in G$ be any (again, anew).
 7. There is $e_1 \in G$ s.t. $\quad e_1a = a \land be_1 = b$.
 8. There is $e_2 \in G$ s.t. $\quad e_2b = b \land ae_2 = a$.
 9. $e_1 = e_2$ because:
$\quad$ 9.1. There is $x \in G$ s.t. $\quad bx = e_1$.
$\quad$ (Because given $b,e_1 \in G$, there is(are) $x',x \in G$ s.t. $\quad x'b = bx = e_1$.)
$\quad$ 9.2. There is $y \in G$ s.t. $\quad yb = e_2$.
$\quad$ (Because given $b,e_2 \in G$, there is(are) $y,y' \in G$ s.t. $\quad yb = by' = e_2$.)
$\quad$ 9.3. $e_1 = bx \quad$ (by 9.1)
$\quad$ 9.4. $\quad = (e_2b)x \quad$ (by 8)
$\quad$ 9.5. $\quad = e_2(bx) \quad$ (by associativity)
$\quad$ 9.6. $\quad = e_2e_1 \quad$ (by 9.1)
$\quad$ 9.7. $\quad = (yb)e_1 \quad$ (by 9.2)
$\quad$ 9.8. $\quad = y(be_1) \quad$ (by associativity)
$\quad$ 9.9. $\quad = yb \quad$ (by 7)
$\quad$ 9.10. $\quad = e_2 \quad$ (by 9.2)
 10.  $\forall a,b \, \exists e:\, ea = ae = a \land eb = be = b. \quad$ (by 6-9)
 11. $ \exists e \, \forall a: \, ea = ae = a. \quad$ (by (H))
 A: We start with  $\forall a,b \, \exists x,y :\, xa = ay = b.$ (1)
Let $a\in G$.
Using (1) for $a,a$, we have that there are $e_l, e_r$ s.t $e_l a = a e_r = a$.
Then, using your argument in 4. , we can show that $e_r = e_l$ (You have to use (1) two more times, for $a,e_l$ and for $a,e_r$)
Therefore, for each $x\in G$ there is $e_x \in G$ s.t $e_x x = x e_x = x$.
Now, let $b \in G$. 
Using your argument in 4. again, we can show that $e_a = e_b$. (Again, you have to use (1) as necessary)
Therefore, we have that all $e_x$ are equal, and thus we have shown that there is an identity element in $G$.
Finally, to show the inverse axiom, start with (1) for $x, e$, for an arbitrary $x \in G$ and $e$ the identity element we have shown that exists.
You get that there are $x_1, x_2 \in G$ s.t $x_1 x = x x_2 = e$
But $x_1 = x_1 e = x_ 1 x x _2 = x_2$.
A: (This answer is merely a more detailed explanation of a part of Dosidis' answer, the part which explains how to derive the identity axiom. This is written by the one who posted the question and accepted his answer, just in case there are other people as slow as himself.)
We start with $\forall a,b \, \exists x,y:xa=ay=b$. (1)  
Let a $\in G$.
Using (1) for $a,a$, we have that there are $e_l, e_r \in G$ s.t. $e_la = ae_r = a$.
Using (1) for $a, e_l$, we have that there are $x, x' \in G$ s.t. $xa = ax' = e_l$.
Using (1) for $a, e_r$, we have that there are $y', y \in G$ s.t. $y'a = ay = e_r$.
$e_l = xa = x(ae_r) = (xa)e_r = e_le_r = e_l(ay) = (e_la)y = ay = e_r$.  
Therefore, we have $\forall x, \exists e_x: \, e_xx = xe_x = x$. (2)
In other words, for each $x \in G$, there is a local identity (so to speak) $e_x$ for it. 
Now, let $a, b \in G$ anew.
Using (2), there is $e_a \in G$ s.t. $e_aa = ae_a = a$ and there is $e_b \in G$ s.t. $e_bb = be_b = b$.
Using (1) for $b, e_a$, we have that there are $x, x' \in G$ s.t. $xb = bx' = e_a$.
Using (1) for $a, e_b$, we have that there are $y', y \in G$ s.t. $y'a = ay = e_b$.
$e_a = xb = x(be_b) = (xb)e_b = e_ae_b = e_a(ay) = (e_aa)y = ay = e_b$.   
Therefore, we have that all local identities $e_x$ are equal, and thus we have shown that there is a global identity (so to speak) in $G$.
