I am sure that the answer to the question of classifying sharply 3-transitive actions of Lie groups on the 2-sphere is: Only the standard action of $PSL(2,C)$, up to topological conjugation. However, the proof, as I see it, is by an unpleasant case-by-case analysis. Here is the proof of a similar result for group actions on $S^1$ (still unpleasant, but fewer cases to analyze).
Suppose that $G$ is a Lie group which has a 3-transitive action on $S^1$. Then $G$ is homeomorphic to the torus $T^3$ minus the big diagonal, i.e. to $Mob(S^1)$, the full Moebius group of the circle, i.e. the full isometry group of the hyperbolic plane (since the latter also acts 3-transitively on the circle). In particular, $G$ has two connected components, is 3-dimensional and its maximal compact subgroup is isomorphic to $O(2)$. The number of 3-dimensional Lie algebras is very small, especially since you already know that the Lie algebra is "noncompact." From this, you get a classification of connected Lie groups with maximal compact subgroup $U(1)=S^1$:
$PSL(2,R)$ or its finite-sheeted covering. Concisely, the Lie algebra of $G$ is isomorphic to $sl(2,R)$.
$R^2\rtimes S^1$.
$Aff_+(R)\times S^1$.
$Nil_3/{\mathbb Z}$, where $Nil_3$ is the group of strictly upper-triangular 3-by-3 matrices and ${\mathbb Z}$ is a subgroup of the (1-dimensional) center of $Nil_3$.
Here $Aff_+(R)$ is the group of orientation-preserving affine transformations of the real line. This gives us all the possible options for $G^0$, the identity component of $G$.
Next, since $G$ acts on $S^1$ transitively, pick a point $z\in S^1$ and let $G_z$ denote the stabilizer of $z$ in $G$. Then $G/G_z$ is homeomorphic to $S^1$. In the cases 2 and 3, the only options for such $G^0_z$ are the (semi)direct factors $R^2$ and $Aff_+(R)$ respectively. But these subgroups act trivially by the left multiplication on the quotients $G^0/G^0_z$, from which we see that in the cases 2 and 3 the action of $G$ on $S^1$ would have to contain the subgroups $R^2$ and $Aff_+(R)$ respectively in the kernel of the action. Hence, a 3-transitive action is impossible in these two cases.
In the fourth case, one check that $Nil_3/{\mathbb Z}$ contains no 2-dimensional subgroups which have trivial intersection with the (central) compact subgroup $S^1$. This rules out the fourth case.
We conclude that the Lie algebra of $G$ is isomorphic to $sl(2,R)$. Let's finish the proof in the case when $G\cong Mob(S^1)$. As above, pick a point $z\in S^1$ and obtain a homeomorphism $G/G_z\cong S^1$. Thus, $G_z$ is a 2-dimensional subgroup of $G$ which has nontrivial intersection with both connected components of $G$ (since $S^1$ is connected and $G$ is not). Up to conjugation, $PSL(2,R)$ has only one 2-dimensional subgroup, namely one isomorphic to $Aff_+(R)$, the stabilizer of "infinity." Thus, $G_z$ has to be conjugate to the subgroup $Aff(R)< G$. From this, we obtain that the action of $G$ on $S^1$ is equivariantly homeomorphic to the action of $G$ on $G/Aff(R)$ by the left multiplication, which is the standard action of the full Moebius group on the circle.
It remains to exclude the cases when $G^0$ is a nontrivial covering of $PSL(2,R)$. The same analysis as above, shows that the only possible actions are given by left multiplication of $G$ on $G/G_z$, where $G^0_z$ (up to conjugation) is the identity component of the preimage of
$Aff_+(R)< PSL(2,R)$ under the covering map $p: G^0\to PSL(2,R)$. Let us understand the corresponding actions of $G^0$ on $S^1$ assuming that $p$ is an $n$-fold covering map, $n\ge 2$.
To get some idea of what is going on, take first $n=2$, i.e. $G^0=SL(2,R)$. The subgroup $G^0_z$ then (up to conjugation) is the group of upper-triangular matrices with positive diagonal entries. The action of $G^0$ on $G^0/G^0_z$ then is the action of $SL(2,R)$ on the space $\Sigma$ of rays emanating from the origin in the plane. Each hyperbolic element $h$ of $SL(2,R)$ (with positive eigenvalues) has four fixed points in $\Sigma$: Two for each eigenspace. Two of these rays ($a^1, a^2$) are attractive (they correspond to the eigenvalue $>1$) and two ($r^1, r^2$) are repulsive. Now, if $g\in G$ swaps $a^1$ and $r^1$, it will have to conjugate $h$ to $h^{-1}$, hence, $g$ necessarily also swaps $a^2$ and $r^2$. Thus, $g$ cannot swap $a^1$ and $r^1$ while, say, leaving $a^2, r^2$ fixed. In particular, the action of $G$ on $\Sigma$ cannot be 3-transitive. The same argument works if you have $n>2$: Each hyperbolic element of $PSL(2,R)$ has a lift $\tilde{h}\in G^0$ which has
$n$ attractive fixed points $a^i$ and $n$ repulsive fixed points $r^j$. One cannot swap $a^1$ and $r^1$ while leaving, $a^2, r^2$ fixed.
Edit. Here is a sketch of the proof for actions on $S^2$. If $G$ admits a sharply 3-transitive action on $S^2$ then $G$ is homeomorphic to $PSL(2,C)$. In particular, the maximal compact subgroup of $G$ is isomorphic to $SO(3)$. From this, one concludes that $G$ is isomorphic to:
Either $PSL(2,C)$.
Or a semidirect product $H\rtimes SO(3)$, where $H$ is a 3-dimensional simply-connected solvable group (of which there is a continuum).
Since $SO(3)$ has only two real linear 3-dimensional representations (trivial and standard, up to isomorphism), one then sees that in Case 2 we have
2i. Either $G\cong H\times SO(3)$.
2ii. Or $G\cong SE(3)$, the group of orientation-preserving isometries of the Euclidean 3-space $R^3$.
Then, as in the case of the analysis of actions on the circle, the point-stabilizer $G_z$ is 4-dimensional, not containing a conjugate of the maximal compact subgroup. In Case 2, all such subgroups will contain the solvable radical $H$, resulting in non-faithful action of $G$ on $G/G_z$, which is a contradiction. This leaves us only with Case 1. In this case each 4-dimensional subgroup is a conjugate of $Aff(C)$, the group of complex-affine transformations. Hence, the action on $S^2$ is standard.
