Find the properties of an ellipse from 5 points in 3D space

I don't know if this is simpler or not, but I'll propose a more geometric approach. All you need is a good routine to find the intersection point of lines $AB$ and $CD$ from the coordinates of points $A$, $B$, $C$, $D$.

Let $ABCDE$ be the five given points. I'll suppose in the following they are on the same plane. We can use Pascal's theorem to find the line tangent to the ellipse at $A$: if $F$ is the intersection of $AB$ with $CD$, and $G$ is the intersection of $AC$ with $BE$, then the intersection $T_A$ of $FG$ with $DE$ is a point on the tangent at $A$ to the conic $ABCDE$ (see here for a detailed proof).

Repeat the same construction with points $B$ and $C$, to find $T_B$ and $T_C$ on the respective tangents. Let then $H$ be the intersection of $AT_A$ with $BT_B$ and $L$ the intersection of $BT_B$ with $CT_C$. If $M$ and $N$ are the midpoints of segments $AB$ and $BC$, then the center $O$ of the ellipse is the intersection of lines $HM$ and $LN$ (this is due to a another well-wknon theorem: center, midpoint of a chord and intersection of the tangents at the endpoints of the chord are collinear).

Having found the center you can now compute the lengths of two conjugate semi-diameters: $$ \alpha=\sqrt{\overline{OH}\cdot \overline{OM}},\quad \beta={\alpha\cdot\overline{AM}\over\sqrt{\alpha^2-\overline{OM}^2}}, $$ while the angle $\theta$ between them is simply the angle between lines $HM$ and $AM$. If it is possible to make an ellipse pass through the five points then $\beta$ is a real number, but that is not a sufficient condition. One should also find point $K$, the intersection between line $OM$ and the line through $C$ parallel to $AB$: the ellipse can be constructed only if the number $$ \beta'={\alpha\cdot\overline{CK}\over\sqrt{\alpha^2-\overline{OK}^2}} $$ is the same as $\beta$.

You can finally find semi-axes $a$ and $b$ of the ellipse by solving the system: $$ a^2+b^2=\alpha^2+\beta^2\\ ab=\alpha\beta\sin\theta. $$