How useful is the assumption that water doesn't auto-ionize in the following?
First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.



Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem.
You can write the full system of equations describing the aqueous equilibrium:


conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).
electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$
water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$
acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$


You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).

You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, this is indeed a good approximation.