If $\forall a,b \in G, \exists x \in G: a * x = b$ and $\forall a,b \in G, \exists x \in G: x * a = b$ then $G$ is a group

Solution 1:

Let $g$ be any element of $G$ and take $a = b = g$ in assumption 2.
It follows that there exists $e \in G$ such that $g * e = g$.

It follows that $h*g*e = h*g$ for any $h\in G$.
By assumption 3, any element of $G$ can be written as $h*g$ for some $h \in G$.
Thus we have $x*e = x$ for any $x \in G$.

This shows that there exists a right identity $e$ of $G$.
The same proof, with left and right switched, shows that there exists a left identity $e'$ of $G$.

We then have $e' = e'e = e$, which means that $e$ is an identity of $G$.

It remains to show that every element has an inverse.
For any $g \in G$, assumptions 2 and 3 say that there exist $h$ and $h'$ such that $g*h = e = h'*g$.
As $h' = h'*e = h'*g*h = e*h = h$, we see that $h$ is an inverse of $g$.