(1) From the problem, we know that the distance from the center of the moving circle $C$ to the point $(2, 0)$ is equal to its distance to the line $x = -2$.

According to the definition of a parabola, the trajectory of the center of the moving circle $C$ is a parabola with focus at $(2, 0)$ and directrix $x = -2$.

Thus, the equation of the trajectory $E$ of the center of the moving circle is $y^2 = 8x$.

(2) Proof: From the problem, we know that when the slope of the line $AB$ is $0$, it does not satisfy the condition. Therefore, let's assume the equation of the line $AB$ is $x = my + 1$. Solving the system of equations:
$$\begin{cases} x = my + 1, \\ y^2 = 8x, \end{cases}$$
we eliminate $x$ and obtain $y^2 - 8my - 8 = 0$, with $\Delta = 64m^2 + 32 > 0$, which always holds true.

Denote $A(x\_1, y\_1)$ and $B(x\_2, y\_2)$, and $M(-1, t)$. Then we have:
$$y\_1 + y\_2 = 8m, \quad y\_1 \cdot y\_2 = -8, \quad x\_1 + x\_2 = 8m^2 + 2, \quad x\_1 \cdot x\_2 = 1.$$
Now, let's compute the slopes:
$$2k\_MP = 2 \cdot \frac{t}{-1 - 1} = -t,$$
$$k\_MA + k\_MB = \frac{y\_1 - t}{x\_1 + 1} + \frac{y\_2 - t}{x\_2 + 1}$$
$$= \frac{y\_1 x\_2 + y\_2 x\_1 + y\_1 + y\_2 - t(x\_1 + x\_2) - 2t}{x\_1 x\_2 + x\_1 + x\_2 + 1}$$
$$= \frac{\frac{1}{8} y\_1 y\_2(y\_1 + y\_2) + y\_1 + y\_2 - t(x\_1 + x\_2) - 2t}{x\_1 x\_2 + x\_1 + x\_2 + 1}$$
$$= \frac{-t(8m^2 + 4)}{8m^2 + 4} = -t.$$
Hence, $\boxed{k\_MA + k\_MB = 2k\_MP}$.