Diego's answer in the comments above is related to why we would expect any classifiable $C^\ast$-algebra to satisfy $A\cong A\otimes \mathcal Z$:
Since $\mathcal Z$ is separable, nuclear, unital, simple and UCT (the properties of $C^\ast$-algebras we wish to classify by $K$-theory and traces),
tensoring with any other $C^\ast$-algebras with these properties will have all the same properties.
And since $\mathcal Z$ has the same $K$-theory and traces as $\mathbb C$, one doesn't change $K$-theory and traces by tensoring with $\mathcal Z$.
Hence if we expect to have classification by $K$-theory and traces, we would expect $A \cong A \otimes \mathcal Z$ for all $C^\ast$-algebras classified by $K$-theory and traces.
Now, I really want to address a misconception which I think is quite common amongst people not working in the
classification or structure programme of nuclear C*-algebras:
$\mathcal Z$-stability (essentially) has nothing to do with the Jiang-Su algebra $\mathcal Z$!
For instance, the UHF algebras, the irrational rotation algebras, and the Cuntz algebras are all $\mathcal Z$-stable, but they have nothing to do with $\mathcal Z$.
This is similar to how McDuff factors* don't really have anything to do with $\mathcal R$,
or how $\mathcal O_\infty$-stable $C^\ast$-algebras have nothing to do with $\mathcal O_\infty$**
One should consider the very natural and frequently occuring regularity property "$\mathcal Z$-stability" as being equivalent
(by very deep, surprising, and non-constructive theorems!) to "$A \cong A\otimes \mathcal Z$".
Similar to the McDuff property, $\mathcal Z$-stability can be characterised by the (norm-)central sequence algebra $\frac{\prod_{\mathbb N} A}{\bigoplus_{\mathbb N} A} \cap A'$ (for separable unital $A$)
being suitably non-trivial, e.g. by requiring that it contains a unital copy of $Z(2,3)$;
the $C^\ast$-algebra of continuous functions $f\colon [0,1] \to M_2(\mathbb C) \otimes M_3(\mathbb C)$ with $f(0) \in M_2(\mathbb C) \otimes 1$ and $f(1) \in 1 \otimes M_3(\mathbb C)$.
Unfortunately, it would be much too technical to explain exactly how this property is used in the classification thereom.
A key technique is that these "almost central homotopies" coming from $Z(2,3)$ can for instance be used to show that $\mathcal Z$-stable $C^\ast$-algebras have cancellation of full projections by $K_0$, that $K_1(A) = U(A)/U_0(A)$
(this was proved in an unpublished paper by Jiang in the late 90's; the paper can be found on arXiv), and to prove that the Cuntz semigroup is almost unperforated (Rørdam).
$\mathcal Z$-stability turns out to be a very mild property that can be verified in an abundance of examples, such as through the Toms-Winter conjecture in the cases where this is known to hold, such as separable, simple, non-type I $C^\ast$-algebras with finite nuclear dimension.
In fact, it is very hard to construct separable simple nuclear non-type I $C^\ast$-algebras which are not $\mathcal Z$-stable (cf. Villadsen, Rørdam, Toms).
So if you have any natural construction that gives you a separable, nuclear, simple $C^\ast$-algebra, you are almost certain to obtain a $\mathcal Z$-stable $C^\ast$-algebra
unless you tried very very hard to construct one without this property.
(*) McDuff factor: separable $II_1$-factors such that $M \cong M\overline{\otimes} \mathcal R$. Equivalently, their ($W^\ast$-)central cequence algebras are non-abelian.
(**) For separable, nuclear, simple $C^\ast$-algebras being $\mathcal O_\infty$-stable is equivalent to being purely infinite by a theorem of Kirchberg.
In particular, the Cuntz algebras $\mathcal O_n$ all satisfy $\mathcal O_n \cong \mathcal O_n \otimes \mathcal O_\infty$ (this is quite spectacular; you can't express this isomorphism explicitly!).
Being $\mathcal O_\infty$-stable is equivalent to the central sequence algebra $\frac{\prod_{\mathbb N} A}{\bigoplus_{\mathbb N} A} \cap A'$ (for separable unital $A$)
containing two isometries $s_1,s_2$ such that $s_1^\ast s_2 = 0$.
And this is the property which is used for classifying purely infinite $C^\ast$-algebras by "moving things around" in a paradoxical way.
