Galileo said $$v_3 = v_1 \oplus_{G} v_2 = v_1 + v_2$$ so $$v \oplus_{G} (-v) = 0$$ and so $$ \frac{c}{2} \oplus_{G} \frac{c}{2} = c$$
Define $$G_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ then the Galiean transform for a space-time interval $$\begin{pmatrix} c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = G_x(v)\begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ \Delta x - v \Delta t \\ \Delta y \\ \Delta z \end{pmatrix} $$ .
Note that $$G_x(0) = I$$ which is the identity transform.
So we wish to solve $$ G_x(v_2) G_x(v_1) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = G_x(v_3) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} $$ for all space-time intervals.
Obviously $$ G_x(v_2) G_x(v_1) = G_x(v_3)$$
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_1}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_2}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_3}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
Expanding the left half, we have :
$$\left( I - \frac{v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) \left( I - \frac{v_1}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \frac{v_1 v_2}{c^2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}^2 = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = G_x(v_1 + v_2) = G_x(v_3)$$.
So $$v_3 = v_1 + v_2$$ as Galileo and Newton would expect.
Einstein said $$v_3 = v_1 \oplus_{E} v_2 = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}$$ so $$v \oplus_{E} (-v) = 0$$ and so $$ \frac{c}{2} \oplus_{E} \frac{c}{2} = \frac{4}{5} c$$
Define $$\Lambda_x(v) = \begin{pmatrix} \cosh \, \tanh^{-1} \frac{v}{c} & \quad & - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & \cosh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}$$ Note that $$\Lambda_x(0) = I$$.
We wish to solve $$\Lambda_x(v_2)\Lambda_x(v_1) = \Lambda_x(v_3)$$
and since $$\begin{pmatrix}\cosh a & \quad & \sinh a \\ \sinh a & \quad & \cosh a \end{pmatrix} \begin{pmatrix}\cosh b & \quad & \sinh b \\ \sinh b & \quad & \cosh b \end{pmatrix} = \begin{pmatrix}\cosh a \cosh b + \sinh a \sinh b & \quad & \cosh a \sinh b + \sinh a \cosh b \\ \cosh a \sinh b + \sinh a \cosh b & \quad & \cosh a \cosh b + \sinh a \sinh b \end{pmatrix} = \begin{pmatrix}\cosh (a + b) & \quad & \sinh (a + b) \\ \sinh (a + b) & \quad & \cosh (a + b) \end{pmatrix} $$,
we have the solution $$v_3 = c \tanh \left( \tanh^{-1} \frac{v_1}{c} \; + \; \tanh^{-1} \frac{v_2}{c} \right) = c \frac{ \frac{v_1}{c} + \frac{v_2}{c} }{1 + \frac{v_1}{c} \frac{v_2}{c}} = \frac{ v_1 + v_2 }{1 + \frac{v_1 v_2}{c^2}}$$.
Masterov says $$v \oplus_{M} (-v) = \cdots$$ ?
Masterov says $$\frac{c}{2} \oplus_{M} \frac{c}{2} = \cdots$$ ?
In general, there is no velocity parameter which makes the Masterov transformation on the right side equal to the product of the two Masterov transformations on the left side.
We don't know what Masterov's equivalent of $$\Lambda_x(v)$$ because he only defines: $$L' = L ( 1 - v^2/c^2) \\ H' = H \sqrt{ 1 - v^2/c^2} \\ T' = T$$ so we
can define:
$$\mathcal{L}_x(v) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -\frac{v A(v)}{c} & 1 - \frac{v^2}{c^2} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ such that
$$\begin{pmatrix} 0 \\ \Delta x' \\ 0 \\ 0 \end{pmatrix} = \mathcal{L}_x(v)\begin{pmatrix} 0 \\ \Delta x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ (1 - \frac{v^2}{c^2}) \Delta x \\ 0 \\ 0 \end{pmatrix} $$
This suggests that $$ 1 - \frac{v_3^2}{c^2} = (1 - \frac{v_2^2}{c^2}) (1 - \frac{v_1^2}{c^2})$$ or $$v_3^2 = v_1^2 + v_2^2 - \frac{v_1^2 + v_2^2}{c^2}$$
But to solve $$v_3 = 0$$ we have the relation $$v_2 = \pm \frac{v_1}{\sqrt{\frac{v_1^2}{c^2} -1}}$$ which if $$0 < \left| v_1 \right| < c$$ requires $$v_2$$ to be
imaginary.
Likewise we can define:
$$\mathcal{H}_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) & -\sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} & 0 \\ 0 & 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos B(v) & - \sin B(v) \\ 0 & 0 &\sin B(v) & \cos B(v) \end{pmatrix}$$ such that
$$\begin{pmatrix} c \Delta t' \\ 0 \\ \Delta y' \\ \Delta z' \end{pmatrix} = \mathcal{H}_x(v)\begin{pmatrix} c \Delta t \\ 0 \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ 0 \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta y \cos B(v) - \Delta z \sin B(v) \right) \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta z \cos B(v) + \Delta y \sin B(v) \right) \end{pmatrix} $$
Similar problems exist, and the physical requirement that v be a observed velocity indicates this cannot describe the physics of this universe.