Suppose $E$ is a three-dimensional Euclidean vector space. The reflection plane $F$ through the origin can be defined by a vector $v$ which is orthogonal to the plane. The reflection of a point $x$ about this plane is the linear isometry of $E$. $$s_Fx = x-2\frac{\langle v,x \rangle}{|v|^2}v$$ Now I want to show that $-s_F$ is the reflection in a line orthogonal to that plane, or similarly the rotation of $180^\circ$ around that axis.
I can easily do this if $F$ is the plane orthogonal to a base vector of E. Since then for a unit vector $e_i$ orthogonal to $F$ and belonging to the basis of E: $s_Fx = x - 2\langle e_i, x \rangle e_i = x-2x_i$ and thus $-s_Fx = -x + 2x_i$. And thus the different kind of reflections can be easily shown.
Now I was wondering if I could choose a similar approach to an arbitrary $F$ one that isn't necessarily orthogonal to one of the base vectors of $E$. I guess taking any change of basis to a basis triple including the the orthogonal vector could let me give a similar argument.

Yes, you are free to change to another orthonormal basis with $f_1,f_2\in F$.
Geometrically, observe that $x\mapsto -x$ is the composition of reflections through the planes $(e_1,e_2),\ (e_1,e_3),\ (e_2,e_3)$ (in any order) for any orthogonal basis $e_1,e_2,e_3$, and the composition of any two of these is the reflection through their intersection.
Thus, with the basis $f_1,f_2,f_3$ above, we get $$-s_F=\left(s_{(f_2,f_3)}\circ s_{(f_1,f_3)}\circ s_{(f_1,f_2)}\right)\circ s_F=s_{(f_2,f_3)}\circ s_{(f_1,f_3)}=s_{f_3}\,. $$