We are given a line $l$. The line is mapped onto itself through a series of projections that involve other lines and -- importantly! -- conics. In the end, points $A$, $B$, and $C$ on $l$ appear to be mapped to itself. Is the composition of projections the identity?
I prove that it is, but I'm not sure about the proof. Could someone verify, please?
Proof. I prove by choosing an arbitrary point $O$ on line $l$. The cross-ratio of $O,A,B,C$ is then given by $\frac{OB}{AB}\cdot\frac{AC}{OC}$. Now assume the transformation sends $O$ to some point $O'$. Since we work only with lines and conics, the cross-ratio is preserved, so we have: $$ \frac{O'B}{AB}\cdot\frac{AC}{O'C}=\frac{OB}{AB}\cdot\frac{AC}{OC}, $$ or $$ \frac{O'B}{O'C}=\frac{OB}{OC}. $$ Now one needs to examine several cases of relative positions of points $O,O',B,C$, also keeping in mind that the distances are signed. This is straightforward and in all of them either $O'O=0$ or it doesn't have a solution, which proves the claim. End of proof.
Would that be good enough?
Yes, you proof seems to be correct, provided you mean what I think you mean with "projections involving lines and conics". I assume you mean projections onto another line w.r.t. a point (not lying on any of those two lines) and stereographic projections onto conics. Also note that the first kind of projection can be seen as a special case of the second kind (using a degenerate conic consisting of two lines).
You could also argue this differently: A bijection $f$ on a line which preserves cross-ratios (in the real case it even is sufficient for $f$ to map harmonic quadruplets to harmonic quadruplets) must be a projective transformation. And a projective transformation on a line is uniquely determined by the image of three points, so if $f$ has three fixed points it must be the identity.