Nice problem! My solution: Invert at the contact point of $S$ and plane $ABCD$, say $X$. From equal tangents from each of the points $A,B,C,D$ separately, we have that $P$ is on a sphere associated with each point (say $S_a$). In fact, since the sphere intersects at a circle with another sphere, the points lie on a circle. Denote by $X_{LMN}$, the image of the tangency point of face $LMN$ with $S$. The points $X_{ABC'} \equiv X_{ABD'}, X_{ACB'} \equiv X_{ACD'}, X_{ADB'} \equiv X_{ADC'}$ are collinear. Thus we have 4 such lines. Let's dropped the primed points from the name, as they are redundant. The circle through $X_{BC}, X_{BD}, X_{CD}$ is the image of the circle intercepted by the tangent cone from $A'$ to $S$, by $S$. Consider the Miquel point of the four lines mentioned. The image of this point is on the sphere, and the line $A'P$ is tangent to the sphere $S$. By symmetry, the lines $B'P, C'P, D'P$ are also tangent to $S$, and thus we have the conclusion from the fact that all tangents at a point on the sphere lie on a plane.

Take the polarity of the problem wrt. $S$. Then the four complanar point became $4$ planes that cut the the sphere in $4$ circles ($w_a,w_b,w_c,w_d$) passing through the tangency point P of the sphere, then the point $A'$ become the plane passing throught the intersection of $w_b,w_c,w_d$ different from $P$ and call the circle in which cuts the sphere $w_{a'}$; do the same construction for the other 3 point $B', C', D'$. Now the condition becomes $w_{a'}, w_{b'},w_{c'},w_{d'}$ all intersect in a common point. Now invert in P with arbitrarius radius, the the problem becomes given 4 lines in a plane $l_a,l_b,l_c,l_d$, inversion of the four circle passing through $P$, then taken circumcircle of the 3 intersection of $l_a,l_b,l_c$ and and simlarly for the other 3 choises of 3 lines, they pass through a single point. But this is just the definition of the miquel point of the quadrilateral, hence we are done. Q. E. D.