In fact it's easy, and hopefully enlightening, to give a direct proof.   Here's one, using the theory of orbifolds (which happens to be how I think about it). (NB I suspect it's not how Fine and Rosenberger think about it.)

Your triangle group is the fundamental group of an orbifold $O$ with underlying space a 2-sphere and cone points of order $\alpha,\beta,\gamma$.  This orbifold has Euler characteristic

$\chi(O)= 2-(1-1/\alpha)-(1-1/\beta)-(1-1/\gamma)=1/\alpha+1\beta+1/\gamma-1$.

For convenience, we will give the proof in the hyperbolic, ie Fuchsian, case where $\chi(O)<0$; a similar proof can be given in the Euclidean ($\chi(O)=0$) case.  It seems clear that the statement is false in the spherical (ie $\chi(O)>0$) case, since then $G$ is finite.  For more information about orbifolds, see Peter Scott's survey article `The geometries of 3-manifolds'.

The orbifold $O$ is the quotient of the hyperbolic plane $\mathbb{H}^2$ by your triangle group $G$.  That is to say, $G$ acts properly discontinuously and cocompactly, but not freely, on $\mathbb{H}^2$.  The cone points on $O$ are precisely the images of the points in $\mathbb{H}^2$ with non-trivial stabilizers.  Each cone point on $O$ has a preimage in $\mathbb{H}^2$ whose stabilizer is generated by one of the generators $a,b$ or $c$.  Call these preimages $x_a,x_b,x_c$ respectively.  

Now suppose that $g\in G$ has finite order.  By the classification of isometries of $\mathbb{H}^2$, it follows that $g$ fixes a point $y$ in $\mathbb{H}^2$.  Thus $y$ has non-trivial stabilizer, and so for some $h\in G$, $y=hx_a$ or $hx_b$ or $hx_c$; wlog, let's say $y=hx_a$.  Therefore $h^{-1}gh$ stabilizers $x_a$, and so $g$ is conjugate into $\mathrm{Stab}_G(x_a)=\langle a\rangle$, as required.