Here is another justification of the answer that there is a pencil of ellipses satisfying your conditions, via a naive dimension count. I will assume that you are interested in the plane $\mathbb{R}^2$ (as opposed to a projective plane, complex plane $\mathbb{C}^2$, etc.). There is a $5$-dimensional family of ellipses in the plane (because there is a $6$-dimensional space of quadratic equations, $a_1 x^2 + a_2 xy + a_3 y^2 + a_4 x + a_5 y + a_6 = 0$, and we lose one dimension to scaling, i.e., multiplying both sides of the equation by the same number). (To be a little more precise these are conic curves. The ellipses are an open subset, defined by a discriminant inequality.) To have a focus at a particular point imposes $2$ conditions. To meet a given ellipse at a particular point is $1$ condition; to meet a given ellipse at a particular point, and be tangent at that point, is $2$ conditions; to meet a given ellipse and be tangent to it at *some* point is $1$ condition. Putting it together, to have a fixed focus and be tangent to two given ellipses is $2+1+1=4$ conditions, on a $5$-dimensional family of ellipses, leaving a $1$-dimensional family of ellipses that meet your conditions.

The only potentially interesting thought I can add is that the fixed focus doesn't have to be related in any way to two given ellipses of tangency. In the above argument it was just an arbitrary point.

Finally, this naive count assumes some "general position." If, for example, the two given ellipses coincide, or if one is degenerate, then some of the $2+1+1$ conditions are redundant, and more ellipses meet your conditions.

Edit: This is NOT a proof. I don't show that there are any real solutions. Even if there are real solutions, I'm just producing a $1$-dimensional family of conic sections; they may not be ellipses.