Your first condition $\mathcal O_ve_1 = v$ is equivalent to saying the map $\mathcal O : S^{n-1} \to SO_n$ is a section of the fibre bundle $$SO_{n-1} \to SO_n \to S^{n-1}$$ If you had a section of that bundle and if it were continuous, then the total space would be a product $$SO_n \simeq SO_{n-1} \times S^{n-1}$$ (as a space). Generally this isn't true, in fact, it's true if and only if the tangent bundle to $S^{n-1}$ is trivial, which is only when the sphere is a Lie group, which is when $n=2$ or $4$. Of course, if you were to puncture the sphere $S^{n-1}$ a continuous section would exist, you can write down explicit sections using linear algebra constructions and a little spherical geometry.

The secondary condition amounts to saying that the commutator $[\mathcal O_v,\mathcal O_w]$ fixes $e_1$ for all $v$ and $w$. This is again too much to expect, even over the punctured sphere. Checking that you can't do this in general amounts to looking at the local properties of the conjugation map on the lie group $SO_n$. The map $A \longmapsto A^{-1}BA$ is an automorphism of the Lie group $SO_n$, so it preserves the identity element, to $SO_n$ acts on the tangent space to the Lie group (called the associated Lie algebra), taking the derivative you get the adjoint map, which is a Lie bracket. This Lie bracket is quite non-degenerate so you can't expect what you're asking for. Anyhow, that's just a sketch. Hope it's useful.

Your first condition $\mathcal O_ve_1 = v$ is equivalent to saying the map $\mathcal O : S^{n-1} \to SO_n$ is a section of the fibre bundle $$SO_{n-1} \to SO_n \to S^{n-1}$$ If you had a section of that bundle and if it were continuous, then the total space would be a product $$SO_n \simeq SO_{n-1} \times S^{n-1}$$ (as a space). Generally this isn't true, in fact, it's true if and only if the tangent bundle to $S^{n-1}$ is trivial, which is when $n=2, 4$ or $8$. Of course, if you were to puncture the sphere $S^{n-1}$ a continuous section would exist, you can write down explicit sections using linear algebra constructions and a little spherical geometry.

The secondary condition amounts to saying that the commutator $[\mathcal O_v,\mathcal O_w]$ fixes $e_1$ for all $v$ and $w$. This is again too much to expect, even over the punctured sphere. Checking that you can't do this in general amounts to looking at the local properties of the conjugation map on the lie group $SO_n$. The map $A \longmapsto A^{-1}BA$ is an automorphism of the Lie group $SO_n$, so it preserves the identity element, to $SO_n$ acts on the tangent space to the Lie group (called the associated Lie algebra), taking the derivative you get the adjoint map, which is a Lie bracket. This Lie bracket is quite non-degenerate so you can't expect what you're asking for. Anyhow, that's just a sketch. Hope it's useful.