 Solution

The continuous functional calculus for a Hermitian element of a C-star algebra identifies its generated unital C-star algebra with continuous functions on its real spectrum, preserves the involution and norm, and sends functions to the corresponding spectral values by spectral mapping. Consequently an element is a square of a Hermitian element of a C-star algebra exactly when it is a Hermitian element of a C-star algebra with nonnegative spectrum: the reverse direction uses the continuous square root on that spectrum.

Nonnegative scalar multiples preserve this property directly. For positive $a,b$ put $C=\Vert a\Vert +\Vert b\Vert$. Functional calculus gives $\Vert \Vert a\Vert 1-a\Vert \le \Vert a\Vert$ and similarly for $b$, so $\Vert C1-(a+b)\Vert \le C$. The sum is a Hermitian element of a C-star algebra, and spectral mapping now places its spectrum in $[0,2C]$. It therefore has a positive square root in a C-star algebra. If $C=0$ both elements are zero. Thus $A_{+}$ is a cone. It is pointed because an element positive with its negative has spectrum contained in $\{0\}$ and hence is zero by the self-adjoint norm formula. It is closed as well: for a converging sequence of Positive elements of a C-star algebra use a common norm bound $C$ and pass to the limit in $\Vert C1-a_n\Vert \le C$.

For the two-by-two matrix the exact criterion isA positive matrix is Hermitian with nonnegative eigenvalues; evaluating its quadratic form at the coordinate vectors gives the diagonal conditions, and its determinant, the eigenvalue product, gives the final inequality. Conversely if $a_{11}>0$, completing the square writes the quadratic form as $a_{11}|z_1+(a_{12}/a_{11})z_2{|}^2+(a_{22}-|a_{12}{|}^2/a_{11})|z_2{|}^2$, which is nonnegative. If $a_{11}=0$, the determinant inequality forces $a_{12}=0$ and the conclusion is immediate. The spectral theorem then gives nonnegative eigenvalues and their Hermitian square root, proving positivity in the question's sense.