A semigroup in which both the equations ax=b and ya=b have unique solution, is a group. Prove it. What is actual meaning of that $ax=b$ and $ya=b$ have unique solution?
Do x and y depend on each other?
Let $G$ be the semigroup
Since $ax=b$ have unique solution in $G$
∴ $\exists\, e\in G$ such that $ae=e \,\,,\forall a\in G$
( Hence $e$ is the right identity of $G$, is it not?)
Similarly, $ya=b$ has unique solution in $G$
∴ $\exists\, f \in G$ such that $fa=a \,\,,\forall a \in G$ 
( Hence $f$ is left identity of $G$, is it not?)
Now  $fe=f$ ( ∵e is the right identity )
and  $fe=e$ (∵f is left identity )
$\implies e=f$
∴ $e$ is the identity of element in $G$   
Is my proof right?
 A: 
What is actual meaning of that ax=b and ya=b have unique solution?

It means that, for every $a$ and $b$ there exists exactly one $x$ such that $ax=b$, and there exists exactly one $y$ such that $ya=b$.
As for your proof: you started well. For every $a$ there will be unique $e$ such that $ae=a$. However, is that $e$ the same for all $a$’s? 
Let’s pick $e$ for one particular $a$ and check that $be=b$  for all other $b$. Now, you know that $b=ya$ for some $y$. Thus, $be=yae=ya=b$.
The rest of the proof looks ok. You may want to complete it (existence of inverse etc) but it is fairly obvious how it will go.
A: This is wrong. You did not justify how you jumped from “for each $a$ and each $b$ there is one and only one $x$ such that $ax=b$” to “there is a $e$ such $(\forall a\in G):ae=e$”. That's a big jump.
A: Given G is a semigroup. So it possesses closure property and associativity as well. Now If we prove 
1- G has a right identity.
2- Every element of G has a right inverse with respect to right identity. Then G will be a group.
For(1)- let a∈G be any.Suppose ax=a, ∃ e in G such that ae=e(since ax=b and ya=b has a solution in G).Now we"ll prove that this e is the right identity in G.
Consider g belongs to G be any, then ya=g has solution in G ⇒ there exist some h in G such that ha=g. Now ge=(ha)e=h(ae)=ha=g⇒  e is the right identity ∀g∈G.
For(2)- let a∈G be any. So ax=e has a solution in G say a'.Then aa'=e so that a has right inverse a' in G.
