These are deep waters.
Central symmetries are defined on the first page of
T. Banakh, A. Dudko, D. Repovš
http://new.math.uiuc.edu/math402/public/models/htranslation.html
http://en.wikipedia.org/wiki/Square_root_of_a_matrix#Square_roots_of_positive_operators
http://en.wikipedia.org/wiki/Bol_loop
Using the Poincare disk model, a hyperbolic translation is the Möbius transformation given by matrix
$$ A \; = \;
\left( \begin{array}{rr}
1 & \alpha \\
\bar{\alpha} & 1
\end{array}
\right) ,
$$
or
$$ T_\alpha(z) = \frac{z + \alpha}{\bar{\alpha}z + 1 } \; \; , $$
where $\alpha$ is a complex number with $|\alpha| < 1.$ The transformation does take the unit circle to itself as along as $\alpha$ is not itself on the unit circle. However, we also need $T_\alpha(0)$ to be inside the disk, and $T_\alpha(0) = \alpha.$
The matrix for $T_\alpha \circ T_\beta$ has matrix
$$
\left( \begin{array}{rr}
1 + \alpha \bar{\beta} & \alpha + \beta \\
\bar{\alpha} + \bar{\beta} & 1 + \bar{\alpha} \beta
\end{array}
\right) .
$$
This is simply not Hermitian, and other things go wrong.
Given two positive Hermitian matrices,
$$ A \; = \;
\left( \begin{array}{rr}
1 & \alpha \\
\bar{\alpha} & 1
\end{array}
\right) ,
$$
and
$$ B \; = \;
\left( \begin{array}{rr}
1 & \beta \\
\bar{\beta} & 1
\end{array}
\right) ,
$$
as you see above $AB$ or $BA$ are not Hermitian, so the classical Bruck loop operation is
$$ \sqrt{B A^2 B},$$ where the square root is the (unique) positive Hermitian matrix $H$ such that $ H^2 = B A^2 B.$ So, the operation is not commutative or associative, but it satisfies a Bol identity and behaves well as far as inverses.
EDIT, Saturday, March 24, 12:23 Pacific time: got it, very hard. We get, with the above $A,B$ as written,
$$ B A^2 B = \left( \begin{array}{rr}
| 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2 & 2 (\alpha + \beta) (1 + \bar{\alpha} \beta) \\
2 (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta}) & | 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2
\end{array}
\right). $$
Now, as matrices, we get
$$ \sqrt{B A^2 B} = \left( \begin{array}{cc}
| 1 + \bar{\alpha} \beta | & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |} \\
\frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |} & | 1 + \bar{\alpha} \beta |
\end{array}
\right). $$
However, as a Möbius transformation, we may divide all four entries by the real number $ | 1 + \bar{\alpha} \beta |,$ and call that transformation $A * B.$ That is,
$$ A * B = \left( \begin{array}{cc}
1 & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |^2} \\
\frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |^2} & 1
\end{array} \right)
$$
or
$$ A * B = \left( \begin{array}{cc}
1 & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ ( 1 + \alpha \bar{\beta} ) ( 1 + \bar{\alpha} \beta ) } \\
\frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{( 1 + \bar{\alpha} \beta ) ( 1 + \alpha \bar{\beta} ) } & 1
\end{array} \right)
$$
or
$$ A * B = \left( \begin{array}{cc}
1 & \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} } \\
\frac{ \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta } & 1
\end{array} \right).
$$
We finally have what Grishkov and Nagy write, using Möbius transformations with both diagonal elements equal to $1$ and the matrix Hermitian in any case, everything is determined by a complex number of modulus less than $1$ in the upper right corner. Put together, we have
$$ A \; = \;
\left( \begin{array}{rr}
1 & \alpha \\
\bar{\alpha} & 1
\end{array}
\right) ,
$$
and
$$ B \; = \;
\left( \begin{array}{rr}
1 & \beta \\
\bar{\beta} & 1
\end{array}
\right) ,
$$
then
$$ A * B = \left( \begin{array}{cc}
1 & \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} } \\
\frac{ \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta } & 1
\end{array} \right).
$$
With $x= \alpha, y = \beta,$ we see the Grishkov and Nagy formula at the bottom of page 6,
$$ x \cdot y = \frac{x + y}{1 + x \bar{y}} = \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} }.$$