I want to create a seam for an ellipse. (Mathematically I think this means that there are constant-length normal lines between the ellipse and the curve that creates the seam, but I'm not 100% sure that definition is accurate.) I know I can't do this by creating another ellipse with the axes reduced by the seam width because the distance between the two ellipses isn't constant. (i.e. in this image the green line is shorter than the two red lines.)
I can convert ellipses into cubic Beziers.
Is there a way to calculate a modification to the control points of the inner Bezier to make the distance between the outer ellipse and inner Bezier constant?


What you've described is typically called an "offset curve". The offset curve for an ellipse is ... not nice. The points of the ellipse itself satisfy a nice quadratic like $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ while those of the offset curve satisfy...a polynomial of degree 8 perhaps? I can't recall, but it's certainly not quadratic, hence not an ellipse (as you observe). But more important, it's also not described by a cubic spline, so that approach won't work either. I'm sorry to say that you just have to just do the calculus and work things out:
At the point $(x, y)$ of the ellipse above, a normal vector is $(\frac{2x}{a^2}, \frac{2y}{b^2})$, so the unit normal is $$ n(x, y) = \frac{1}{\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4}}}(\frac{x}{a^2}, \frac{y}{b^2}) $$ and your offset curve is at location $$ (x, y) + s n(x, y), $$ where $s$ is the offset distance (positive for "larger than the original ellipse"; negative for "smaller than").
That only works for relatively small values of negative $s$; for larger negative numbers, you get "focal points" and things get messy.