Another proof makes use of the pigeonhole principle. Fix $a \in G$, and consider the map
$$
f : G \to G, \qquad x \mapsto a x.
$$
Since $G$ is cancellative, the map is injective, hence surjective because $G$ is finite, so that there is $b \in G$ such that $e = f(b) = a b$.
Similarly, there is $c \in G$ such that $c a = e$.
Thus $$c = c e = c (a b) = (c a) b = e b = b,$$ and $b = c$ is the required inverse of $a$.
You need not assume that there is an identity, but can prove that one exists.
First show, with the arguments above, that any element $z$ of $G$ can be written in the form $z = x a$, for some $x \in G$. Then show that there is $s \in G$ such that $a s = a$. Now $$z s = x a s = x a = z, \quad\text{for all $z \in G$,}$$ so $s$ is a right identity. Similarly there is a left identity $t$ such that $$t z = z,\quad \text{for all $z \in G$,}$$ and finally $$t = t s = s$$ is the identity.
A semigroup with a left identity and unique right inverses is not necessarily a group. Here is an example:
Let $G$ be any set with more than one element and call one of the elements $e$. Define a multiplication on $G$ by $x\cdot y = y$. This is associative, since $x\cdot(y\cdot z) = (x\cdot y)\cdot z = z$. For all $x\in G$ we have $e\cdot x = x$, so $e$ is a left identity (as is every other element of $G$). Also, $x\cdot y = e$ if and only if $y = e$, so each $x$ has a unique right inverse, namely $e$. But $(G,\cdot)$ is not a group, since there is no right identity.
EDIT: Looking closely at your axiom $3$, you seem to be requiring something more than unique right inverses. It seems that you also want an element to be the right inverse of at most one element; equivalently, if an element has a left inverse, it must be unique. In this case, $(G, \cdot)$ must indeed be a group.
To prove this, it suffices to show that a right inverse is also a left inverse. Choose any $x\in G$ and consider the product $x'\cdot x\cdot x'\cdot x''$, where $x'$ denotes a right inverse. On the one hand, $$x'\cdot (x\cdot x')\cdot x'' = x'\cdot e \cdot x'' = x'\cdot x'' = e.$$ On the other, $x'\cdot x\cdot (x'\cdot x'') = x'\cdot x\cdot e$, which must also be $e$. Since the right inverse of $x'$ is unique, $x'' = x\cdot e$. It follows that
$$
x''\cdot x' = (x\cdot e)\cdot x'= x\cdot(e\cdot x') = x\cdot x' = e.
$$
Hence $x'$ is a right inverse for both $x$ and $x''$, so $x = x''$. Finally, $x'\cdot x = x'\cdot x'' = e$, so the right inverse of $x$ is also its left inverse.
Best Answer
First note that once you found that $e_aa=a$ you can multiply both sides by $a$ from the right side and get $(ae_a)a=a^2$. By the cancellation rule you get $ae_a=a$. Now take any $b\in G$. You know there is an element $x\in G$ such that $xa=b$. Multiply both sides by $e_a$ on the left side and you will get:
$be_a=xae_a=x(ae_a)=xa=b$
Now you know that $be_a=b$. Multiply both sides by $b$ from the left side to get $b(e_ab)=b^2$ and by the cancellation rule you get that also $e_ab=b$.