Here is another way to prove the proposition. Curves will be called $c$ and $d$ for short. I will assume (without real loss of generality) that $(a,b)=(-1,1)$. Furthermore, upon switching (if needed) to the pair of curves $(c,-d)$, we can assume $\beta_1=\beta_2$ and call this function simply $\beta$. As a last simplification, you can further assume $c(0)=d(0)$ and $c^.(0)=d^.(0)$ : just switch to the couple of curves $(c, P\cdot d)$ for the appropriate rotation $P$.
Since $c$ and $c^.$ are curves on the sphere, you get upon differentiation that $ c\perp c^.,~c^.\perp c^{..}$ and $\langle c, c^{..}\rangle=-1$. Consider now the basis $C(t)=(c(t),c^.(t),c(t)\wedge c^.(t))$ : this is a direct orthonormal basis of your space, and thus the determinant defining $\beta$ can be calculated in this basis. Also what precedes shows that $c^{..}=-c+(\cdots)c\wedge c^.$ which because of the determinant calculations is actually $c^{..}=-c+\beta\cdot c\wedge c^{.}.$ Exactly the same holds for $d$.
We also set $(e,f,g)=(c(0),c^.(0),c(0)\wedge c^.(0)=C(0)=D(0)$.
Now consider the family of rotations $R_c(t)$ defined by sending the vectors of $C(0)$ to those of $C(t)$ in order of appearance : by definition, for all $t$, we have $c(t)=R_c(t)\cdot c(0)$. Now differentiate this family with respect to $t$ : the resulting map has $\frac{dR_c}{dt}(t):\lbrace \begin{array}{ll} e\mapsto & c^.(t) \\ f\mapsto & c^{..}(t)=-c(t)+\beta(t)c(t)\wedge c^.(t) \\ g\mapsto & c(t)\wedge c^{..}(t)=\beta(t) c(t)\wedge(c(t)\wedge c^.(t))=-\beta(t)c^.(t) \end{array}$ In other words, $\frac{dR_c(t)}{dt}=R_c(t)\times \tau(t)$ where $\mathrm{Mat}(\tau(t);(e,f,g))= \left(\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & -\beta(t) \\ 0 & \beta(t) & 0 \end{array}\right)$
Exactly the same equations stand if we do things with the analoguoulsy defined $R_d(t)$. We can now show that $R_c=R_d$ and thus for all $t,~c(t)=R_c(t)\cdot e=R_d(t)\cdot e=d(t)$. You can either say that $R_c(0)=R_d(0)=\mathrm{id}$ and they are both solutions to the same first order linear differential equation, and by the Cauchy Lipschitz theorem they (are globally defined and) coincide, or you can differentiate $R_c(t)(R_d(t))^{-1}=R_c(t)R_d(t)^*$ by hand and get $\frac{dR_c(t)R_d(t)^*}{dt}=R_c(t)\tau(t)\times R_d(t)^*+R_c(t)\times (-\tau(t)R_d(t)^*)=0,$ where we have made use of the skew symmetry of $\tau(t)$ this shows equality for all $t$.