Relative lowering of vapour pressure
In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property.

A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature dependence of formulation (assuming ideal solutions) by making the measurement relative to another quantity, which in this case is the vapor pressure of the pure solution. Otherwise, the influences on vapor pressure are multifactorial. I.e. external pressure may affect it, as well as temperature.

Here is a somewhat hand-wavy out-of-the-pocket calculation to demonstrate. Assume the vapor pressure can be expressed as an ideal gas. Therefore pressure may be expressed using the ideal gas law, $P = \frac{nRT}{V}$. Also know Raoult's law for vapor pressure, $P_s = \chi_s P_o$ where $\chi_s$ is the mole fraction. Using your definitions, we have for absolute pressure difference
$$P_o - P_s = P_o - \chi_s P_o = \left(1-\chi_s\right)P_o = \left(1-\chi_s\right)\frac{n_oRT}{V}$$
This is therefore dependent upon temperature. Now do the same for the relative difference in pressures
$$1-\frac{P_s}{P_o} = 1 - \frac{\chi_sP_o}{P_o} = 1-\chi_s$$
The expression is now dependent upon the moles only, i.e. a colligative property. Of course, in reality this depends on the assumptions we made under ideal Raoult conditions.