Existence of universal enveloping inverse semigroup (similar to "Grothendieck group") Context
In its simplest form, the Grothendieck group construction associates an abelian group to a commutative semigroup in a "universal way".
Now I'm interested in the following nilpotent commutative semigroup $N$ consisting of two elements $a$ and $b$ such that $a^2=b^2=ab=ba=a$. The corresponding Grothendieck group is the trivial group with just one element, which is a bit boring. So I asked myself whether it would be possible to construct a "universal enveloping inverse semigroup"(*) in a similar way as the Grothendieck group, and whether it would be more interesting.
I tried to compute the "universal enveloping inverse semigroup" $N_I$ for the semigroup $N=\{a,b\}$. I got $N_I = \{ a, b, b^{-1}, bb^{-1}, b^{-1}b \}$ with $(b^{-1})^2=ab^{-1}=b^{-1}a=a$. What I find surprising is that $N_I$ is not commutative, even so $N$ is commutative. So I tried to compute the "universal enveloping commutative inverse semigroup" $N_C$ instead and got $N_C = \{ a \}$.
(*)Note: In a semigroup $S$, we say that $y\in S$ is an inverse element of $x\in S$, if $xyx=x$ and $yxy=y$. A semigroup $S$ is called an inverse semigroup, if each $x\in S$ has a unique inverse element $x^{-1}\in S$. It's easy to see that $xx^{-1}$ and $x^{-1}x$ are idempotent, that all idempotent elements in an inverse semigroup commute, and that $(xy)^{-1}=y^{-1}x^{-1}$. So at least superficially, inverse semigroups seem to be nice generalization of groups and can have a zero element without being trivial.
Question
Does the "universal enveloping inverse semigroup" (and the "universal enveloping commutative inverse semigroup") of a semigroup $S$ always exist? I guess the answer is yes and this probably follows from some theorem of universal-algebra. Similarly, I guess that the "universal enveloping regular semigroup" doesn't always exist and wonder whether this also follows from some theorem of universal-algebra.

I now found out how to prove that no "universal enveloping regular semigroup" exists for the example given in the question. (The existence of the other two cases has already been proved in the answer by Martin Wanvik.)
Let 
$a=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, 
$b=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, 
$b'=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, 
$c=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$, 
$c'=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$, 
$d=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$, 
$d'=\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$, 
$e=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, and 
$f=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. Let $N := \{ a,b \}$ and $N_I := \{ a,b,b',e,f \}$ (as in the question with $b'=b^{-1}$, $e=bb^{-1}$ $f=b^{-1}b$). It is easy to see that $S := \{ a,b,b',c,c',d,d',e,f \}$ is a regular group, and that $N_I$ and $R := \{ a,b,c,d,e \}$ are regular sub-semigroups of $S$. Now $b^{-1}=b'$ in $N_I$ and $c$ is the unique inverse element of $b$ in $R$. So if there would exist a regular semigroup $N_R$ containing $N$ as sub-semigroup, for which it is possible to uniquely extend any homomorphism from $N$ to $N_I$ or $R$, then the extension to $N_R$ of a homomorphism $h$ from $N$ to $S$ (with $h(a)=a$ and $h(b)=b$) won't be unique.
