$f(x)=rx-{x\over1+x}$. Fixed points where $f(x)=x$, so $x=rx-{x\over1+x}$ $x^*=0$ or $1=r-{1\over1+x},\quad {1\over1+x}=r-1,\quad 1+x={1\over r-1},$ $x^*={1\over r-1}-1={2-r\over r-1}$ I don't know how you got $x^*={1\over r}-1$.
Bifurcations occur when $f'(x^*)=\pm1$. When $f'(x^*)=-1$, you may get a period-doubling bifurcation, but that's not among the ones you listed, so I'll ignore that. $f'(x)=r-{1\over(x+1)^2}$ so we need $r-{1\over(x+1)^2}=1$. When $x^*=0$, this gives $r=2$. The other fixed point, $x^*={2-r\over r-1}$, gives nothing (but check my algebra here).
So, bifurcation occurs at $(r^*,x^*)=(2,0)$ Just from drawing the graphs of $x=0$ and $x={2-r\over r-1}$ and seeing how they cross at $(2,0)$ is, I think, enough to see that it's a transcritical bifurcation. But you can always calculate $f'(x)$ for fixed points nearby $(2,0)$ to see where you have stable fixed points and where unstable, and analyze the bifurcation that way. Or, you can use the formulas in Elaydi, for example, if ${\partial f\over\partial x}(r^*,x^*)=1,\quad {\partial f\over\partial r}(r^*,x^*)=0,\quad {\rm and}\quad {\partial^2 f\over\partial x^2}(r^*,x^*)\ne0$ it's transcritical.