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.