Indeed we have to show that the set is closed under addition in each case. For example, consider $2/3$ and $1/3$. Both of them satisfy $|.|<1$ but $|2/3+1/3|$ is not strictly smaller than $1$. So c doesn't make a groupid either. The similar story can be seen for d for example. In fact $|3/2+(-4/2)|<1$ while $|3/2|\ge1,~~|-4/2| \ge1$.
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
We write $xy$ for $x*y$.
Let $a$ be an element of $G$, and consider the function $f:G \to G$, $f(g)=ag$. This is injective by (2), and so (since $G$ is finite) it must also be surjective. Thus there exists an element $e$ of $G$ such that $f(e)=a$, i.e. $a=ae$.
We want to show that $e$ is in fact a two-sided identity for $G$. So let $b$ be an arbitrary element of $G$, and consider $ab=(ae)b=a(eb)$. Thus by (2), $b=eb$ for all $b$, i.e. $e$ is a left identity. To show $e$ is also a right identity, consider $bb=b(eb)=(be)b$, and apply (3) to conclude that $b=be$.
Now we need to show the existence of inverses. As above, for every $a$ in $G$ the function $g \mapsto ag$ is surjective, so there is an element, $a_R$ say, such that $aa_R=e$. Similarly, there is an element $a_L$ such that $a_La=e$. Since $a_L=a_Le=a_L(aa_R)=(a_La)a_R=ea_R=a_R$, we have that $a_L=a_R$ is a two-sided inverse for $a$.