There are infinitely many ellipses, even assuming that the two tangent lines intersect.
This follows from the very nice property of an ellipse:
If line $t$ is a tangent to an ellipse at point $P$, and if the focii of the ellipse are $A$ and $B$, then the ray $\overrightarrow{BP}$ reflected on $t$ passes through $A$.
Basically, if you shoot a laser from $B$ to any point on the ellipse, and the ellipse is a perfect mirror, then the reflected ray will pass through $A$!
This can be proven using the physics principle: light takes the route which takes the shortest time
. Using the fact that locus of an ellipse is $PA + PB = $ constant, if light from $B$ reflected on the ellipse has to take the shortest time, it can be shown that it has to pass through $A$.
This gives us a way of generating multiple ellipses.
Suppose the given points are $P$ and $Q$ and the tangents ($p$ and $q$ respectively) intersect at $O$, also assume that $\angle{POQ}$ is not acute.
The two tangents divide the plane into four regions. Consider the region $R$ which contains both $P$ and $Q$ on its border.
Now reflect $P$ on the line $q$ at point $Q$. This line intersects $R$ and we get a half-line $S_P$. Similarly reflect $Q$ on the line $p$ at $P$, and we get half-line $S_Q$.
Now given a point $B_i$ (candidate for a focus), we pick a corresponding point $A_i$ (candidate for the other focus) as follows (based on the above reflection property):
Shoot lasers from $B_i$ to $P$ and $Q$ (assuming lines $p$ and $q$ are mirrors) and if they intersect, we pick the point of intersection to be $A_i$.
If it turns out that $PB_i + PA_i = QB_i + QA_i$, then we have our ellipse.
Now given a point $B_1$ on $S_Q$ the corresponding $A_1$ we get is the point $Q$ itself.
We have that $B_1Q + A_1Q \lt B_1P + A_1P$
For a point $B_2$ on $S_P$, the corresponding $A_2$ we get is the point $P$ itself.
We have that $B_2Q + A_2Q \gt B_2P + A_2P$
Now if we pick $B_1$ and $B_2$ sufficiently far from $P$ and $Q$, as we move from $B_1$ to $B_2$, along the segment $B_1 B_2$, there will be a point $B_3$ for which $B_3Q + A_3Q = B_3P + A_3P$ (by continuity arguments) and giving us an ellipse we need.
Now we can pick multiple lines parallel to $B_1B_2$ and use that to get multiple ellipses (at most two parallel lines can give the same ellipse).
If $\angle{POQ}$ is acute, we can pick $B_1$ and $B_2$ on $p$ and $q$ itself, with $P$ lying between $O$ and $B_1$ and $Q$ lying between $O$ and $B_2$ and $B_1$, $B_2$ being sufficiently far from $O$.