The classification of equilibria is intended for isolated equilibria. Since we have a line of non-isolated equilibria, the classification does not fully apply. We must proceed with caution...
Seeing that the trace of the Jacobian matrix is $ax-b$, we can conclude that the equilibria are stable when $x and unstable when $x>b/a$. There is no conclusion at $x=b/a$, because adding a nonlinear term to a center can make it an unstable equilibrium.
Fortunately, for this system we can find the equations of trajectories very easily. Divide the second equation by the first to get $\frac{dy}{dx}=-1+\frac{b}{ax}\tag1$ Integration yields $y=-x+\frac{b}{a}\ln |x| +C \tag2$ as equation of trajectories. One has to be careful in interpreting (2), because trajectories do not actually cross $y=0$: they take infinite time approaching it. If $C$ is such that the curve given by (2) crosses $y=0$, that curve is the union of three trajectories: one above the axis, and two below it.
Consider the case $b/a>0$, when the questionable point falls into the halfplane $x>0$. All trajectories (2) are vertical translates of the same curve, which is concave with maximum at $x=b/a$. Taking $C$ such that the curve is just barely below the $y=0$ line, we conclude that the equilibrium at $x=b/a$ is unstable: there are unbounded trajectories that begin arbitrarily close to it. Here's a picture drawn with Sage for the case $a=b=1$.

When $b/a<0$, the conclusion is the same: instability at $x=b/a$, but now the relevant unbounded trajectories lie above the line $y=0$.
When $b=0$, we still have instability at $x=0$, which is easier to see since the system becomes essentially one-dimensional ($x+y$ is constant).
All in all, this is pretty hard homework for a typical ODE class.