Velocity control system is stable even though CLTF has a pole at s=0?
The control system tries to make the error signal \$r(t) - b(t)\$ zero. Now note that the feedback element \$H(s)\$ has a zero at the origin. This means that, if the output of the plant is a position signal, then \$\frac{10}{s+5}\$ smooths it with a low pass filter and the \$s\$ then differentiates it. So \$b(t)\$ has units of velocity. This means that the input \$r(t)\$ also has the units of velocity.
Since the command is in units of velocity and the output has units of position, to find the (steady-state) velocity-error, we need to look at \$r(t) - \frac{dy(t)}{dt}\$. The differentiation of the output introduces a zero which cancels the pole at origin. So if we looked only at the velocities in the system, we will not see a pole at the origin and so the system will appear stable.
The position signal still grows unbounded for a step input; but we just happen to define our error as velocity error which is bounded.
Alternate explanation

Since the plant has two poles at origin and the feedback has one zero at origin, we can redraw the system so that the velocity state is separated out from the position state. Now, looking only at the velocity control system, if we redo the analysis, it can be seen to be stable with finite steady state error.