I haven't checked the details of this myself, so I can't tell you the correct answers to your questions, but I would suggest that you try to apply the adjoint functor theorem of category theory:
http://en.wikipedia.org/wiki/Adjoint_functors#General_existence_theorem
To translate the problem into categorical language, let $\mathbf{Smgrp}$ denote the category of semigroups (with semigroup homomorphisms as morphisms) and let $\mathbf{Inv}$ denote the category of inverse semigroups (the morphisms are, again, semigroup homomorphisms; it is easy to see that a morphism of semigroups preserve inverses as you've defined them above). Clearly, $\mathbf{Inv}$ is a subcategory of $\mathbf{Smgrp}$, so there is a forgetful functor $F: \mathbf{Inv} \to \mathbf{Smgrp}$.
Since the construction here is called the universal enveloping inverse semigroup of a semigroup, we should expect that the construction constitutes a left adjoint functor to $F$.
To spell out the universal property, this means that if $S_{I}$ is the universal inverse semigroup of the semigroup $S$, then there is a semigroup homomorphism $f: S \to S_{I}$ (formally, a morphism $f: S \to F(S_{I})$) such that whenever $g: S \to H$ ($g: S \to F(H)$ is a semigroup homomorphism with $H$ an inverse semigroup, then there exists a unique semigroup homomorphism $h: S_{I} \to H$ such that $g = h \circ f$ (formally, $g = F(h) \circ f$).
It is now a simple matter of checking that the conditions of the theorem are satisfied; the category $\mathbf{Inv}$ is complete (it has products and equalizers, and hence all (small) limits), so the theorem applies. If you want the semigroups involved to be commutative, simply change the categories appropriately.
More details
Proof sketch that the left adjoint exists:
First of all, we should look at limits in $\mathbf{Smgrp}$ and $\mathbf{Inv}$; they turn out to be "the same": if $\{ S_{i} \}_{i \in I}$ is some family of semigroups, the product is simply the cartesian product of the underlying sets, with the obvious "pointwise" operation. If all semigroups are inverse semigroups, the result is an inverse semigroup. You can easily check that it has the requisite universal property for a product, see
http://en.wikipedia.org/wiki/Product_(category_theory)
If $f,g: S_1 \to S_2$ are semigroup homomorphisms, their equalizer is
$$E = \{ s \in S_1 \,|\, f(s) = g(s) \}$$
together with the inclusion homomorphism $e: E \hookrightarrow S_1$. $E$ is a semigroup in the obvious way, and is an inverse semigroup if $S_1$ and $S_2$ are (if $x \in E$ have inverse $y \in S_1$, we have $f(x) = g(x)$, and it follows that both $g(y)$ and $f(y)$ is the inverse of $f(x) = g(x)$; by uniqueness, $f(y) = g(y)$ so $y \in E$ also). You can again check that $(E,e)$ has the requisite universal property, see
http://en.wikipedia.org/wiki/Equalizer_(mathematics)
Hence both $\mathbf{Smgrp}$ and $\mathbf{Inv}$ are (small-) complete categories, see
http://en.wikipedia.org/wiki/Limit_(category_theory)
(under Existence of limits).
We get that $F$ preserves limits for free, since limits are constructed in exactly the same manner in both categories, and the morphisms are the same. What remains is the solution set condition. Fix a semigroup $S$, and consider a semigroup homomorphism $f: S \to F(H)$, for some inverse semigroup $H$. The idea is to take the solution set to be the isomorphism classes of inverse semigroups generated by $f(S)$ for some $f$. This will be a set if the cardinality of the inverse semigroup $\langle f(s) \rangle \subseteq H$, generated by $f(S) \subseteq H$ is bounded, for a given $S$. This appears to be the case here, since you can start with $G = f(S) \cup f(S)^{-1}$ (where $f(S)^{-1}$ denotes the set of inverses in $H$ for elements in $f(S)$), and consider all finite products of elements in $G$. This will be an inverse semigroup (because, as you mentioned, the formula $(xy)^{-1} = y^{-1}x^{-1}$ holds). You still need to check that this actually is a solution set, but that should be reasonably easy.
Note that this only answers the question of existence, it doesn't provide an explicit construction. My guess is that coming up with one isn't going to be too difficult (and it appears that you already have some ideas in that direction, at least for specific examples), perhaps the construction of the Grothendieck group can be emulated.