I think we only need $ax=b$ or $ya=b$ have solutions in $G$, I'll prove it.
Proof:
(I)
$G$ is a group $\implies$ $a^{-1}ax=a^{-1}b$ $\implies$ $x=a^{-1}b$ $\implies$ $ax=b$ has solutions in $G$
(II)
(A)
$ax=b$ has solutions in $G$ $\implies$ $ax=a$ has solutions in $G$ $\implies$ there exists an identity element $e∈G$ such that $ae=ea=a$
(B)
$ax=b$ has solutions in $G$ $\implies$ $ax=e$ has solutions in $G$ $\implies$ there exists an inverse element $a^{-1}∈G$ such that $ae=ea=a$
Hence, $ax=b$ has solutions in the semigroup $G$ (for all $a,b\in G$) if and only if $G$ is a group.
Correct?
Reference: Fraleigh p. 49 Question 4.39 in A First Course in Abstract Algebra
Best Answer
As has been pointed out in comments, there are gaps in the proof as it stands.
We give a start towards a correct proof. We will need the solvability of both types of equation, $ax=b$ and $ya=b$.
Let $a$ be some fixed element of our semigroup. By what you wrote, there is an object $e_a$ such that $ae_a=a$.
We show that $be_a=b$ for any $b$. To show this, observe that there is a $y$ such that $b=ya$. Multiplying on the right by $\color{darkcyan}{e_a}$, we get $$\begin{align} b\color{darkcyan}{e_a} & = (ya)\color{darkcyan}{e_a} \\ &= y(a\color{darkcyan}{e_a}) \\ &= ya \\&= b \end{align}$$
This shows that our semigroup has a right identity.
In a similar way, one can show that the semigroup has a left identity.
It is quite easy to show that if $e_r$ is a right identity and $e_l$ is a left identity then $e_r=e_l$.
Showing the existence of the inverse of any $b$ is left to you.