= Solution
Write a matrix in the set as $M(a,b)$. The zero matrix $M(0,0)$ has no multiplicative <inverse matrix>, so $S$ is \b[not a group]. Its <determinant> is
$$
\det M(a,b)=|a|^2+|b|^2,
$$
which vanishes only at that zero matrix. It is therefore the only element that needs to be removed.
For closure, direct <matrix multiplication> gives
$$
M(a,b)M(c,d)=M(ac-bd^*,\ ad+bc^*).
$$
The product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The <identity matrix> is $M(1,0)$, <associativity> is inherited from matrix multiplication, and
$$
\boxed{M(a,b)^{-1}=\frac1{|a|^2+|b|^2}\begin{pmatrix}a^*&-b\\b^*&a\end{pmatrix}
=M\left(\frac{a^*}{|a|^2+|b|^2},\frac{-b}{|a|^2+|b|^2}\right).}
$$
Thus $\boxed{S\setminus\{0_{2\times2}\}\text{ is a group}}$. It is the multiplicative group of nonzero <quaternions> in the <complex matrix representation of quaternions>.
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