EDIT 1: I just noticed that I had misread the word edges and solved instead for faces. However the proof for edges does not differ much from my solution for faces. EDIT 2: The proof has bugs. What a mind blowing theorem! But unfortunately I proved this only in one direction. I was not able to subdue the converse. Anyway here is my proof that if this type of octahedron is circumscribed around an ellipsoid then its diagonals intersect at a point. Because ellipsoid is the affine image of sphere it suffices to prove this fact for octahedron circumscribed around a sphere. My proof hardly relies on the following theorem. Theorem. If two cones are tangent to a sphere then their intersection is unity of two conics. Proof. Denote by $A, B$ the vertices of the cones and by $I$ the center of the sphere ($\Omega$). Let $\omega$ be the section of $\Omega$ with plane $ABI$. Let the tangents from $A$ and $B$ to $\omega$ form the convex quadrilateral $CDEF$ and let $\omega$ touch the sides $DE, EF, FC, CD$ at $K, L, M, N$ respectively. Obviously the cones are tangent to $\Omega$ with circles $\omega_A$ and $\omega_B$ that are perpendicular to $ABI$ and have diameters $LN$ and $MK$. Note that $A$ is the pole of $LN$ wrt $\omega$ and $B$ is the pole of $KM$ so $AB$ is the polar of $X=KM \cap LN$. This means that $Y=LM \cap KN$ lies on $AB$. $YM \cdot YL=YK \cdot YN$ so there exists an inversion with center $Y$ that the circle $\omega_A$ with diameter $LN$ and perpendicular to $ABI$ sends to the circle $\omega_B$ with diameter $MK$ and perpendicular to $ABI$. It follows that any line through $Y$ intersecting $\omega_A$ will intersect $\omega_B$ too. Also note that $ML$ and $KN$ intersect on $CE$ by Menelaus and note that $MK$ and $NL$ intersect on $CE$ by Brianchon, that is the points $C, E, X, Y$ are collinear. Now consider any section of $\Omega$ with plane $\pi$ through $AB$. Let it be the circle $\omega_1$. As above construct tangents to $\omega_1$ intersecting at pints $C_1, D_1, E_1, F_1$ and label the respective contact points of $\omega$ with $K_1, L_1, M_1, N_1$. For the same reason as mentioned above the lines $L_1N_1$ and $K_1N_1$ intersect on $AB$. Clearly they intersect at $Y$. Also the point $X_1=M_1K_1 \cap N_1L_1$ lies on the intersection line of planes of $\omega_A$ and $\omega_B$ that is perpendicular to $ABI$ due to symmetry. On the other hand again by the reason mentioned above points $C_1, E_1, X_1, Y_1$ are collinear. It follows that the points $C_1, E_1$ lie in a plane passing through $CE$ and perpendicular to $ABI$. Thus as $\pi$ rotates around $AB$ the points $C_1, E_1$ of intersection of two cones run with plane curve which must be a conic. Analogously the other set of common points of two cones (namely the one which contains $D_1, E_1$) lie on a different conic as needed. The converse is not hard to prove with continuity. $\square$ Now back to our problem. Denote our octahedron $ABCDEF$ so that $AD, BE, CF$ are its diagonals. Let $\Omega$ be its insphere. Let $\kappa_1$ and $\kappa_2$ be the conics of intersection of cones with vertices $A$ and $D$ tangent to $\Omega$. Let $BC$ be tangent to $\kappa$ at $T$ where $\kappa$ is one of the conics $\kappa_1$ and $\kappa_2$. Both the planes $ABC$ and $DBC$ intersect the plane of $\kappa$ with lines tangent to $\kappa$ at $T$ which is unique. Thus the planes $ABC$ and $DBC$ intersect by a line lying in the plane of $\kappa$. This means that $BC$ lies in the plane of $\kappa$. Note that each one from the segments $BC, CE, EF, FB$ is tangent to either $\kappa_1$ or $\kappa_2$. If some opposite pair from segments $BC, CE, EF, FB$ is tangent to the same conic ($\kappa_1$ or $\kappa_2$) then they lie in the plane of the same conic thus the diagonals $BE$ and $CF$ intersect. Suppose that $\kappa_1$ is tangent to some adjacent pair from segments $BC, CE, EF, FB$ say $BC$ and $CE$. The last two segments lie in the plane of $\kappa_1$. Draw the second tangents to $\kappa_1$ from $B$ and $E$ and mark the point $F'$ of their intersection. $F'$ lies in the all four planes $ABF, DBF, AEF, DEF$ and so coincides with $F$. Thus in this case too the points $B, C, E, F$ lie in a plane and the diagonals $BE$ and $CF$ intersect. Thus in all cases the diagonals $BE$ and $CF$ intersect. Analogously diagonals $AD$ and $BE$, $AD$ and $CF$ also intersect and because these three diagonals lie in different planes they all have a common point as needed.

Invert in \(T_{AB}\). Keep notations the same. Then \(\gamma_{A}, \gamma_{B}\) become parallel lines and \(\gamma_D, \gamma_E\) become circles tangent to each other and each tangent to one of \(\gamma_A, \gamma_B\). Then \(T_{DE}\) is their internal homothety center and hence the common tangent \(t\) through \(T_{DE}\) to \(\gamma_D, \gamma_E\) is parallel to \(\gamma_A, \gamma_B\). This means that before inversion the preimage of \(t\) was a circle through \(T_{AB}, T_{DE}\) tangent to the four circles \(\gamma\). Hence the circles \(t, \gamma_A, \gamma_B\) had a radical center, i.e. a point \(Z\) for which the segments \(ZT_{AB}, ZT_{BD}, ZT_{DE}\) are equal and tangent to \(\Gamma\). But this means that \(Z\equiv Y\) as needed.

The problem could be solved by other methods too, at least the "only if" part: (a) Using the fact that the intersection of two cones touching a sphere lies in two planes or (b) Inverting in \(T_{EF}\), for example and getting a planar construction with two parallel lines \(\gamma_{E}, \gamma_F\) and many circles, which would turn out to be a symmetric diagram.