(1) The equations of lines $AB$, $AC$ and $BC$ are $y = \frac{4}{3}(x+1)$, $y = -\frac{4}{3}(x-1)$ and $y = 0$ respectively. The distances from point $P$ to $AB$, $AC$ and $BC$ are respectively
$$d_1 = \frac{1}{5} | 4x - 3y + 4 |,$$
$$d_2 = \frac{1}{5} | 4x + 3y - 4 |, \quad d_3 = | y |.$$
According to the assumption, $d_1 d_2 = d_3^2$, we have $|16x^2 - (3y-4)^2| = 25y^2$. That is
$$16x^2 - (3y-4)^2 + 25y^2 = 0, \text{ or } 16x^2 - (3y-4)^2 - 25y^2 = 0.$$
By simplifying the above equations, we obtain that the locus equations of point $P$ consist of
$$\text{circle S: } 2x^2 + 2y^2 + 3y - 2 = 0$$
and
$$\text{hyperbola T: } 8x^2 - 17y^2 + 12y - 8 = 0.$$

(2) The incenter of $\triangle ABC$ is also a point satisfying the assumption.
In view of $d_1 = d_2 = d_3$, solving the equations we have $D(0, \frac{1}{2})$.
Line $L$ passes through $D$, and has three common points with the locus of point $P$. So the slope of $L$ is defined. Suppose that the equation of $L$ is
$$y = kx + \frac{1}{2}.$$
(i) If $k=0$, then $L$ is tangent to the circle $S$, which means there is a unique common point $D$. In this case line $L$ is parallel to the $x$-axis, which implies that $L$ and hyperbola $T$ have two other common points different from point $D$. Hence, there are just three common points for $L$ and the locus of point $P$.

Case 1: Line $L$ passes through point $B$ or point $C$, which means that the slope of $L$ is $k = \pm \frac{1}{2}$, and the equation of $L$ is $x = \pm (2y-1)$. Substitute it into equation $\textcircled{2}$ we get
$$y(3y-4) = 0.$$
Solving it we have $E(\frac{5}{3}, \frac{4}{3})$ or $F(-\frac{5}{3}, \frac{4}{3})$, which means that line $BD$ and curve $T$ have 2 intersection points $B$ and $E$, and line $CD$ and curve $T$ have 2 intersection points $C$ and $F$.

Case 2: Line $L$ does not pass through point $B$ and point $C$ (i.e. $k \neq \pm \frac{1}{2}$). Since for $L$ and $S$ there are two different intersection points, there exists a unique common point for $L$ and hyperbola $T$. Thus for the following system of equations
$$\begin{cases} 8x^2 - 17y^2 + 12y - 8 = 0, \\ y = kx + \frac{1}{2}, \end{cases}$$
there is one and only one real solution. After eliminating $y$ and simplifying we have
$$(8 - 17k^2)x^2 - 5kx - \frac{25}{4} = 0.$$
The above equation has a unique real solution if and only if
$$8 - 17k^2 = 0$$
or
$$(-5k)^2 + 4(8 - 17k^2) \frac{25}{4} = 0.$$
Solving $8 - 17k^2 = 0$ we get $k = \pm \frac{2\sqrt{34}}{17}$. And solving $(-5k)^2 + 4(8 - 17k^2) \frac{25}{4} = 0$ we obtain $k = \pm \frac{\sqrt{2}}{2}$.
Consequently, the set of all possible values of the slope $k$ of line $L$ is the following finite set:
$$\{0, \pm\frac{1}{2}, \pm\frac{2\sqrt{34}}{17}, \pm\frac{\sqrt{2}}{2}\}$$