It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
A subgroup is a subset of a group that is itself closed under the group operation.
A semigroup is a set equipped with an operation that is merely associative, different from a group in that we assume the binary operation of a group is associative and invertible, i.e. each element has an inverse with respect to the operation.
Best Answer
Briefly, the answer is no: associative quasigroups are always inverse semigroups, but inverse semigroups are not necessarily associative quasigroups. However, associative quasigroups are equivalently cancellative inverse semigroups!
To be concrete, I'll work with the following definitions:
On the other hand:
So the question is: are they the same thing? If $(Q,*)$ is an associative quasigroup and $x\in Q$, then we want to see if it has an inverse $y$ such that $x=x*y*x$ and $y=y*x*y$. What would be a candidate? By the Latin square property with $a=b=x$, there is a unique $l$ such that $l*x=x$. Then by the Latin square property again with $a=x$ and $b=l$, there is a unique $r$ such that $x*r=l$. Therefore, $x=l*x=x*r*x$, so $y=r$ is a candidate for an inverse!
Does it work? First, we need to see if $r*x*r=r$, which might not look apparently true, but it has to do with the fact that quasigroups have the cancellation property. In this particular example, we reason as follows:
First note that $(l*l)*x=l*(l*x)=l*x=x$. Since $l$ is the unique element with the property that $l*x=x$, this means $l*l=l$.
Now, this means $x*(r*x*r)=(x*r)*(x*r)=l*l=l$. However, $r$ is the unique element such that $x*r=l$, so again we get to conclude from $x*(r*x*r)=x*r$ that $r*x*r=r$, as desired!
Now we need to know that $r$ is the unique inverse of $x$. Therefore, suppose $y\in Q$ is another inverse of $x$, so that $x=x*y*x$ and $y=y*x*y$. We want to conclude that $y=r$. We can once again use the cancellation property:
Conclusion: If $Q$ is an associative quasigroup, then it is also an invertible semigroup.
Now for the converse: suppose $S$ is an invertible semigroup, then will it necessarily be an associative quasigroup? Unfortunately, the answer is no, and a counterexample is the following: let $S = \{0,1\}$ with the multiplication operation given by the usual multiplication of numbers.
However, $S$ is not an associative quasigroup: try the Latin square property with $a=0$ and $b=1$. We would need to find some $r\in S$ such that $a\cdot r=b$. That is, we are trying to solve $0\cdot r=1$. The left hand side, however, is always zero, so no such $r$ can exist, and therefore $S$ fails to satisfy the Latin square property!
Conclusion: An inverse semigroup is not necessarily an associative quasigroup.
The reason for this failure ultimately boils down to the fact that inverse semigroups don't necessarily have the cancellation property like quasigroups do. If we further assert that our inverse semigroup is cancellative, then we are in luck: for any $a,b\in S$, we can take $l := b\cdot a^{-1}$ and $r := a^{-1}\cdot b$ so that $a\cdot r=b$ and $l\cdot a=b$.
I'll show why this works for $r$. First, uniqueness is clear: if $a\cdot r=b$ and also $a\cdot r'=b$, then by the cancellation property, $a\cdot r=a\cdot r'$ will force $r=r'$. Now to check that $a\cdot r=b$. To do so, we will again use the cancellation property
Conclusion: Associative quasigroups are the same as cancellative inverse semigroups; that is, $(A,\bullet)$ is an associative quasigroup if and only if it is a cancellative inverse semigroup