The same maximal distance occurs in each quadrant, so we can restrict attention to $t\in[0,\pi/2]$. The tangent vector at $t$ is $(-a\sin t,b\cos t)$. This vector is normal to the line, so we just have to take the scalar product of a unit vector in this direction with the position vector in order to find the distance of the origin from the line:
$ \begin{eqnarray} D &=& \left\lvert\frac{(-a\sin t,b\cos t)}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\cdot(a\cos t,b\sin t)\right\rvert \\ &=& \frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\;. \end{eqnarray}$
Differentiating with respect to $t$ yields
$\frac{a^2\sin^4 t-b^2\cos^4t}{\left(a^2\sin^2t+b^2\cos^2t\right)^{3/2}}\;,$
and setting this to zero yields
$a^2\sin^4t=b^2\cos^4t\;,$
$t=\arctan\sqrt{\frac ba}\;.$
Using $\cos t=1/\sqrt{1+\tan^2 t}$, we can evaluate $D$ at this parameter:
$ \begin{eqnarray} D &=& \frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}} \\ &=& \frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\cos t \\ &=& \frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\frac1{\sqrt{1+\tan^2 t}} \\ &=& \frac{(a^2-b^2)\sqrt{b/a}}{\sqrt{a^2(b/a)+b^2}}\frac1{\sqrt{1+b/a}} \\ &=& \frac{a^2-b^2}{a+b}\;. \\ &=& a-b\;. \end{eqnarray}$
The result obviously supports your idea that there might be a simpler way to do this.