I will give an intrinsic characterization below for what this unit class modulo 12th powers means, which may be viewed as an answer of sorts: it expresses the obstruction to extracting the 12th root of a certain canonical isomorphism between 12th powers of line bundles (and so one could shift the answer to: where does the need to extract such a 12th root come up?)

For any ring $R$, the group $R^{\times}/(R^{\times})^{12}$ naturally maps into the degree-1 fppf cohomology of $\mu_{12}$ over ${\rm{Spec}}(R)$, so it classifies isomorphism classes of certain $\mu_{12}$-torsors for the fppf topology over this base. (Namely, those $\mu_{12}$-torsors whose pushout to a $\mathbf{G} _m$-torsor is trivial.)

It is the same to use the etale topology when $12$ is a unit in $R$ (as then $\mu_{12}$ is etale over $R$). So the issue is to associate to any elliptic curve $f:E \rightarrow {\rm{Spec}}(R)$ over a ring a canonical $\mu_{12}$-torsor (with the extra property that its pushout to a $\mathbf{G} _m$-torsor is trivial).

In the theory of Weierstrass planar models for elliptic curves $E$ over a base scheme $S$ (this includes the condition "good reduction") there is an obstruction to the existence of such a model, namely whether or not the line bundle $\omega_{E/S} = f_{\ast}(\Omega^1_{E/S})$ on $S$ admits a global trivialization. The necessity of such triviality is due to the fact that a Weierstrass model produces a trivialization (the ${\rm{dx}}/(2y+\dots)$ thing), and the sufficiency is explained in Chapter 2 of Katz-Mazur (where they use a choice of trivializing section to distinguish some formal parameters along the origin and pass from this to a Weierstrass model via the relationship between global 1-forms, the relative cotangent space ${\rm{Cot}}_e(E)$ along the identity section $e$, and $\mathcal{O}(ne)/\mathcal{O}((n-1)e) \simeq {\rm{Cot}}_e(E)^{ \otimes -n}$ for $n = 2, 3$).

That being said, regardless of whether or not the line bundle $\omega_{E/S}$ is trivial (though it always is when $S$ is local), the line bundle $\omega_{E/S}^{\otimes 12}$ is canonically trivial (in a manner that is compatible with base change and functorial in isomorphisms of elliptic curves): that is the meaning of the classical fact that the product of $\Delta$ with the 12th power of the section ${\rm{d}}x/(2y+\dots)$ is invariant under choice of Weierstrass model. This also underlies Mumford's calculation (recently revisited by Fulton-Olsson) of the Picard group of the moduli stack of elliptic curves as $\mathbf{Z}/12\mathbf{Z}$, which one could regard as providing a distinguished role to that trivialization. Working with the compactified moduli stack over $\mathbf{Z}$ (so allowing generalized elliptic curves with geometrically irreducible but possibly non-smooth fibers, and hence working with relative dualizing sheaf to generalize $\omega_{E/S}$ when allowing non-smooth fibers), the trivialization (which we could generously attribute to Ramanujan) is unique up to a sign, which in turn is nailed down by the Tate curve over $\mathbf{Z}[[q]]$ and the isomorphism of its formal group with $\widehat{\mathbf{G}}_m$. So this trivialization is really a canonical thing, independent of any theory of Weierstrass models.

Letting $\theta_{E/S}$ denote this intrinsic trivializing section of $\omega _{E/S}^{\otimes 12}$ as just defined, it is natural to ask if $\theta _{E/S}$ is the 12th power of a trivializing section of $\omega _{E/S}$. Note that this is a nontrivial condition even when $\omega _{E/S}$ is trivial (such as when $S$ is local). Anyway, the functor of such 12th roots is a $\mu _{12}$-torsor over $S$ for the fppf topology (and etale if 12 is a unit on the base), and as such it corresponds to the inverse of the class of $\Delta$ in the question (for which the base was local). So that is an answer of sorts: it describes the obstruction to extracting a 12th root of the canonical trivialization of $\omega^{\otimes 12}$ obtained by pullback from the trivialization over the moduli space of elliptic curves (up to an issue of signs in the exponent). Now does one ever care to extract such a 12th root? That's another matter...