Need to prove or disprove an alternative group definition. Let be the non empty set $G$ with the operation $*$ satisfy the following 3 conditions:

*

*$a(bc) = (ab)c$, for all $a, b, c \in G$ (associative law).

*For every $a,b$ there is $c$ such that $ac = b$. ($c$  is the "path" from $a$ to $b$).

*For every $a,b$ there is $d$ such that $bd = a$. ($d$ is the "path back" from $b$ to $a$).


Prove or disprove: $G,*$  is a group.

Asociativity:
Proof:
from 1)
Identity element:
Proof:
According to 2) For every $a,a$ there is $e$ such that $ae = a$.
Let proof $ea=a$.
According to 2) For every $e,e$ there is $X$ such that $eX= e$.
Let's check the two possibilites:

*

*$X=a,$

*$X\neq a$
If 1) then $ea=a$ proved.
If 2): $eX=e$ then what?
Every try failed. Someone can do it?

If the third condition is written as

For every $a,b \in G$ there is $d \in G$ such that $db=a$.

then the answer is that $G$ is a group and the existence of the identity element can shown as follows:
Fix $a \in G$. Then there exists $x_a$ and $y_a$ in $G$ such that $ax_a=a=y_aa$. We will prove that $gx_a=g=y_ag$ for any other $g \in G$. Indeed, if $g \in G$, since we know that there exists $x,y \in G$ with $ax=g=ya$, then $$gx_a = (ya)x_a = y(ax_a) = ya = g$$ and similarly $y_ag = g$. In particular $x_a=y_a$ (why?) and then $e := x_a (=y_a)$ is the identity element.
