A slicker proof than this, that we can factor this type of Hermitian matrix in this way I'm going to give my solution to the problem I pose, but my question is, can you solve the problem more elegantly and without referencing another theorem (I build off the finite-dimensional Spectral Theorem, but maybe instead one could do a simplified spin-off on the proof of that). I'd like the steps to be just as elementary as mine below.
Throughout assume everything is projective, i.e. everything lives in $\mathrm{PSL}_2(\mathbb{C})$ where we identify multiples of $-1$.
(I don't actually use this in my proof, but maybe you can!)
Prove: for any $2\times2$ Hermitian matrix $H$ such that $\det(H)=1$,
there exists a matrix $M$ such that $\det(M)=1$
and $H=MM^*$,
where $*$
is the conjugate transpose.
It is well known that $H$ can be diagonalized by unitary matrices (Spectral Theorem). That is, there exist matrices $U,D$ such that $U^*=U^{-1}$, $D$ is diagonal, and $U^*HU=D$.
Write $U=\begin{pmatrix}r & s\\t & u\end{pmatrix}$,
then since $\begin{pmatrix}u & -s\\-t & r\end{pmatrix}
=\begin{pmatrix}\overline{r} & \overline{t}\\\overline{s} & \overline{u}\end{pmatrix}$, we have
$U=\begin{pmatrix}r & s\\ -\overline{s} & \overline{r}\end{pmatrix}$,
and
since $\det(H)=1$, it follows that $\det(D)=\det(U)=1$ so $D$ is of the form $D=\begin{pmatrix}a & 0\\ 0 & a^{-1}\end{pmatrix}$.
Lastly, we can get our desired equation by putting this together and splitting $D$ as follows.
$$  H=\begin{pmatrix}
    r&s\\
    -\overline{s}&\overline{r}
   \end{pmatrix}^*
   \begin{pmatrix}
    a^{-1}&0\\
    0&a
   \end{pmatrix}
   \begin{pmatrix}
    r&s\\
    -\overline{s}&\overline{r}
   \end{pmatrix}\\
  =\begin{pmatrix}
    \overline{r}&-s\\
    \overline{s}&r
   \end{pmatrix}
   \begin{pmatrix}
    \sqrt{a^{-1}}&0\\
    0&\sqrt{a}
   \end{pmatrix}
   \begin{pmatrix}
    \sqrt{a^{-1}}&0\\
    0&\sqrt{a}
   \end{pmatrix}
   \begin{pmatrix}
    r&s\\
    -\overline{s}&\overline{r}
   \end{pmatrix}\\
  =\begin{pmatrix}
    \overline{r}\sqrt{a^{-1}}&-s\sqrt{a}\\
    \overline{s}\sqrt{a^{-1}}&r\sqrt{a}
   \end{pmatrix}
   \begin{pmatrix}
    r\sqrt{a^{-1}}&s\sqrt{a^{-1}}\\
    -\overline{s}\sqrt{a}&\overline{r}\sqrt{a}
   \end{pmatrix}$$
Hermitian matrices have real eigenvalues, so $a\in\mathbb{R}$. Thus $M=\begin{pmatrix}
    \overline{r}\sqrt{a^{-1}}&-s\sqrt{a}\\
    \overline{s}\sqrt{a^{-1}}&r\sqrt{a}
   \end{pmatrix}$ will do the trick.
 A: Your solution looks good. I would be less explicit in order to see the forest more clearly through the trees. Denote the eigenvalues of $H$ by $\lambda_1, \lambda_2 \in \mathbb{R}$ with $\lambda_1 \cdot \lambda_2 = \det(H) = 1$. By replacing $H$ with $-H$ if necessary, we can assume that $\lambda_1, \lambda_2 > 0$. The matrix $H$ is unitary diagonalizable so $H = UDU^{*}$ for $U \in M_2(\mathbb{C})$ unitary and $D = \mathrm{diag}(\lambda_1, \lambda_2)$. 
Define $D' = \mathrm{diag}(\sqrt{\lambda_1}, \sqrt{\lambda_2})$ (here, $\sqrt{\lambda_i}$ is the regular positive square root of $\lambda_i$)  and $M = UD'U^{*}$. Then 
$$\det(M) = \det(D') = \sqrt{\lambda_1}{\sqrt{\lambda_2}} = \sqrt{\lambda_1 \cdot \lambda_2} = \sqrt{1} = 1 $$
and
$$ MM^{*} = UD'U^{*}(UD'U^{*})^{*} = UD'U^{*}(UD'^{*}U^{*}) = UD'^2U^{*} = UDU^{*} = H $$
(where we used $U^{*}U = I$ as $U$ is unitary). In fact, the $M$ constructed above is also Hermitian as
$$ M^{*} = (UD'U^{*})^{*} = UD'^{*}U^{*} = UD'U^{*} = M $$
and $D'$ is diagonal with real values. Thus, $H = MM^{*} = M^2$.
