Letting $T=\pmatrix{S & 0 \cr 0 & S^*}$, you can show that the map
$$
\pi:n\in {\mathbb Z} \mapsto \left\{\matrix{T^n, & \hbox{if } n\geq 0, \cr (T^*)^{-n},& \hbox{otherwise.}}\right.
$$
is a partial representation of the group ${\mathbb Z}$, so your algebra is a quotient of the partial group C*-algebra $C^*_{par}(\mathbb Z)$. Moreover your algebra may be described as the crossed product of its diagonal subalgebra by a partial action of the group $\mathbb Z$, and in this sense it is a universal C*-algebra.

Elaborating a bit more on my answer, the crucial point is to analyze the spectrum of the subalgebra generated by the set $$ \big \{π(n)π(n)^*: n∈ {\mathbb Z}\big \}, $$ which turns out to be homeomorphic to the two-point compactification of ${\mathbb Z}$, namely ${\mathbb Z}∪\{∞, -∞\}$.

In the case of the monogenic inverse semi-group, whose C*-algebra is the same as I called $C^*_{par}({\mathbb Z})$, and is much bigger than the one you described, the spectrum of that abelian subalgebra turns out to be the Cantor set $\{0, 1\}^{\mathbb Z}$.