From what I can see the signed distance you are asking about should not possess a closed form expression. Elliptical coordinates are not a constant distance apart, see http://en.wikipedia.org/wiki/Elliptic_coordinate_system
Evidently this is a difficult computational problem, http://http.developer.nvidia.com/GPUGems3/gpugems3_ch34.html
However, you can get a fair amount of the functionality you need with simple differential geometry, as you do not need to define the ellipsoid as a mesh. You can tell instantly whether a point is outside or inside based on your $f(x,y,z),$ so your comment about $\pm$ sign is entirely correct.
If a point is outside the ellipsoid (f > 0,) you can find the closest point on the ellipsoid by fairly rapid numerical methods. The closest point to some $(r,s,t)$ is the point $(x,y,z)$ in the same octant where the normal vector at $(r,s,t)$ is parallel to the vector $(x,y,z) - (r,s,t).$ The normal vector at $(r,s,t)$ is just a positive multiple of $ \left(\frac{r}{a^2}, \frac{s}{b^2},\frac{t}{c^2}\right).$ So, take as seed value the multiple $(\lambda x,\lambda y, \lambda z).$ At each stage, update the point on the ellipse by projecting the vector $(x,y,z) - (r,s,t)$ onto the tangent plane at $(r,s,t).$ Then, say, project that radially onto the ellipsoid.
I'll need to think about speed. For computer applications, a fast enough algorithm may be good enough, when a closed-form expression is not available.
Inside the ellipsoid, there is not necessarily a unique closest point on the surface. I might recommend just scaling your $f,$ just $- \sqrt{|f|}.$
Meanwhile, I recommended a certain book to Joseph O'Rourke, who does this sort of thing. John Thorpe, Elementary Topics in Differential Geometry.