Let me write a non-geometrical answer.
Your conjecture is false : If $(x_0-u)(y_0-v)=0$, then the line $BC$ does not pass through any fixed point.
I got the following :
If $AB,AC$ are tangent to the circle $(x-x_{0})^2+(y-y_{0})^2=r^2\ (\color{red}{0\lt r\lt \sqrt{(x_0-u)^2+(y_0-v)^2}})$ with $\color{red}{(x_0-u)(y_0-v)\not=0}$, then the line $BC$ passes through a fixed point $(F_x,F_y)$ where
$$\color{red}{F_x=\frac{a^2v(y_0+x_0-u-v)(y_0-x_0+u-v)+u(a^2+b^2)(x_0-u)(y_0-v)}{(a^2-b^2)(x_0-u)(y_0-v)}}$$
$$\color{red}{F_y=\frac{b^2u(y_0+x_0-u-v)(y_0-x_0+u-v)-v(a^2+b^2)(x_0-u)(y_0-v)}{(a^2-b^2)(x_0-u)(y_0-v)}}$$
(Note that the condition $r\lt b$ can be removed and that we have to have $r\lt \sqrt{(x_0-u)^2+(y_0-v)^2}$ in order to draw two tangent lines $AB,AC$. Also, let me emphasize that the circle can exist anywhere as long as it satisfies $0\lt r\lt\sqrt{(x_0-u)^2+(y_0-v)^2}$.)
The equation of the line passing through $A(u,v)$ is given by $mx-y-mu+v=0$.
We want this line to be tangent to the circle, so we have
$$r=\frac{|mx_0-y_0-mu+v|}{\sqrt{m^2+(-1)^2}},$$
i.e.
$$(r^2-x_0^2-u^2+2ux_0)m^2+2(x_0y_0-vx_0-uy_0+uv)m+r^2-y_0^2-v^2+2vy_0=0$$
$$\implies m_1+m_2=\frac{2(-x_0y_0+vx_0+uy_0-uv)}{r^2-x_0^2-u^2+2ux_0},\quad m_1m_2=\frac{r^2-y_0^2-v^2+2vy_0}{r^2-x_0^2-u^2+2ux_0}\tag1$$
Eliminating $y$ from $mx-y-mu+v=0$ and $x^2/a^2+y^2/b^2=1$ gives
$$(b^2+a^2m^2)x^2+(2a^2mv-2a^2m^2u)x-2a^2muv-a^2b^2+a^2m^2u^2+a^2v^2=0$$
Since one of the solutions is $u$, by Vieta's formula,
$$u+X=-\frac{2a^2mv-2a^2m^2u}{b^2+a^2m^2}\implies X=\frac{-2a^2mv+a^2m^2u-ub^2}{b^2+a^2m^2}$$
$$Y=mX-mu+v=\frac{-a^2m^2v-2ub^2m+b^2v}{b^2+a^2m^2}$$
So, we can write
$$B\left(\frac{-2a^2m_1v+a^2m_1^2u-ub^2}{b^2+a^2m_1^2},\frac{-a^2m_1^2v-2ub^2m_1+b^2v}{b^2+a^2m_1^2}\right)$$
$$C\left(\frac{-2a^2m_2v+a^2m_2^2u-ub^2}{b^2+a^2m_2^2},\frac{-a^2m_2^2v-2ub^2m_2+b^2v}{b^2+a^2m_2^2}\right)$$
Using $(1)$, we have
$$b_y-c_y=\frac{-a^2m_1^2v-2ub^2m_1+b^2v}{b^2+a^2m_1^2}-\frac{-a^2m_2^2v-2ub^2m_2+b^2v}{b^2+a^2m_2^2}$$
$$\begin{align}&=\frac{(-a^2m_1^2v-2ub^2m_1+b^2v)(b^2+a^2m_2^2)-(-a^2m_2^2v-2ub^2m_2+b^2v)(b^2+a^2m_1^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2b^2(m_1-m_2)(ua^2m_1m_2-a^2v(m_1+m_2)-ub^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)b^2(ua^2\frac{r^2-y_0^2-v^2+2vy_0}{r^2-x_0^2-u^2+2ux_0}-a^2v\frac{2(-x_0y_0+vx_0+uy_0-uv)}{r^2-x_0^2-u^2+2ux_0}-ub^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)(ub^2(a^2-b^2)r^2+k_1)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}\end{align}$$
where
$$k_1=b^2(b^2u(x_0-u)^2+2a^2v(x_0-u)(y_0-v)-a^2u(y_0-v)^2)$$
and
$$b_x-c_x=\frac{-2a^2m_1v+a^2m_1^2u-ub^2}{b^2+a^2m_1^2}-\frac{-2a^2m_2v+a^2m_2^2u-ub^2}{b^2+a^2m_2^2}$$
$$\begin{align}&=\frac{(-2a^2m_1v+a^2m_1^2u-ub^2)(b^2+a^2m_2^2)-(-2a^2m_2v+a^2m_2^2u-ub^2)(b^2+a^2m_1^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2a^2(m_1-m_2)(-b^2v+ub^2(m_1+m_2)+a^2vm_1m_2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2a^2(m_1-m_2)(-b^2v+ub^2\frac{2(-x_0y_0+vx_0+uy_0-uv)}{r^2-x_0^2-u^2+2ux_0}+a^2v\frac{r^2-y_0^2-v^2+2vy_0}{r^2-x_0^2-u^2+2ux_0})}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)(a^2v(a^2-b^2)r^2+k_2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}\end{align}$$
where
$$k_2=a^2(b^2v(x_0-u)^2-2b^2u(x_0-u)(y_0-v)-a^2v(y_0-v)^2)$$
and
$\quad -c_xb_y+b_xc_y$
$$\small\begin{align}&=\frac{2a^2m_2v-a^2m_2^2u+ub^2}{b^2+a^2m_2^2}\cdot\frac{b^2v-a^2m_1^2v-2ub^2m_1}{b^2+a^2m_1^2}+\frac{a^2m_1^2u-2a^2m_1v-ub^2}{b^2+a^2m_1^2}\cdot\frac{b^2v-a^2m_2^2v-2ub^2m_2}{b^2+a^2m_2^2}\\\\&=\frac{(2a^2m_2v-a^2m_2^2u+ub^2)(-a^2m_1^2v-2ub^2m_1+b^2v)+(-2a^2m_1v+a^2m_1^2u-ub^2)(-a^2m_2^2v-2ub^2m_2+b^2v)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)(-a^4v^2m_1m_2-a^2b^2u^2m_1m_2-b^4u^2-a^2b^2v^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)(-a^4v^2\frac{r^2-y_0^2-v^2+2vy_0}{r^2-x_0^2-u^2+2ux_0}-a^2b^2u^2\frac{r^2-y_0^2-v^2+2vy_0}{r^2-x_0^2-u^2+2ux_0}-b^4u^2-a^2b^2v^2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)}\\\\&=\frac{2(m_1-m_2)(-r^2(a^2+b^2)(a^2v^2+b^2u^2)+k_3)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}\end{align}$$
where
$$k_3=(a^2v^2+b^2u^2)(a^2(y_0-v)^2+b^2(x_0-u)^2)$$
So, the equation of the line $BC$ is given by
$$(b_x-c_x)y=(b_x-c_x)x-c_xb_y+b_xc_y,$$
i.e.
$$\frac{2(m_1-m_2)(a^2v(a^2-b^2)r^2+k_2)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}y=\frac{2(m_1-m_2)(ub^2(a^2-b^2)r^2+k_1)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}x+\frac{2(m_1-m_2)(-r^2(a^2+b^2)(a^2v^2+b^2u^2)+k_3)}{(b^2+a^2m_1^2)(b^2+a^2m_2^2)(r^2-(x_0-u)^2)}$$
which can be written as
$$r^2(k_4y-k_5x+k_6)=-k_2y+k_1x+k_3$$
where
$$k_4=a^2v(a^2-b^2),\quad k_5=ub^2(a^2-b^2),\quad k_6=(a^2+b^2)(a^2v^2+b^2u^2)$$
Solving the system $k_4y-k_5x+k_6=-k_2y+k_1x+k_3=0$ gives that the line $BC$ passes through
$$\left(\frac{k_3k_4+k_2k_6}{k_2k_5-k_1k_4},\frac{k_3k_5+k_1k_6}{k_2k_5-k_1k_4}\right)=\left(\frac{a^2v(y_0+x_0-u-v)(y_0-x_0+u-v)+u(a^2+b^2)(x_0-u)(y_0-v)}{(a^2-b^2)(x_0-u)(y_0-v)},\frac{b^2u(y_0+x_0-u-v)(y_0-x_0+u-v)-v(a^2+b^2)(x_0-u)(y_0-v)}{(a^2-b^2)(x_0-u)(y_0-v)}\right)$$
if $(x_0-u)(y_0-v)\not=0$.
If $x_0-u=0$ and $y_0-v\not=0$, then we have to have
$$k_4y-k_5x+k_6=0\implies a^2vy-ub^2x=\frac{-(a^2+b^2)(a^2v^2+b^2u^2)}{a^2-b^2}$$
but
$$-k_2y+k_1x+k_3=\frac{-2a^2b^2(y_0-v)^2(a^2v^2+b^2u^2)}{a^2-b^2}$$
is not equal to $0$.
If $x_0-u\not=0$ and $y_0-v=0$, then we have to have
$$k_4y-k_5x+k_6=0\implies a^2vy-ub^2x=\frac{-(a^2+b^2)(a^2v^2+b^2u^2)}{a^2-b^2}$$
but
$$-k_2y+k_1x+k_3=\frac{2a^2b^2(x_0-u)^2(a^2v^2+b^2u^2)}{a^2-b^2}$$
is not equal to $0$.
