How much horsepower to weight is required for a 1:1 thrust ratio?
Horsepower cannot be converted into thrust without knowing the speed at which this is done. In your case this is zero, so the special equation for static thrust applies. With $T$ for thrust, $P$ for power, $\rho$ for air density and $d_P$ and $\eta_P$ the propeller diameter and efficiency, respectively, this is $$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{P}^2\cdot\pi\cdot d_P^2\cdot\frac{\rho}{2}}$$
With this equation you cannot simply say how many HP are required to lift that many pounds; instead, you need to add the propeller geometry. If we assume a very efficient propeller with a large diameter, we can combine blade size and mass into disk loading – here 8.5 kg/m² or 41.5 lbs/ft² is a good value. Now we can write, using standard atmospheric density at sea level:
$$\frac{T_0^2}{P^2\cdot\eta_{P}^2} = \frac{2\cdot1.225\,\frac{\text{kg}}{\text{m}^3}}{8.5\,\frac{\text{kg}}{\text{m}^2}\cdot g} = 0.0294\,\frac{\text{s}^2}{\text{m}^2}$$
assuming for simplicity a 1 m propeller diameter. If we now use 70% for propeller efficiency, as you did in the question, the power to mass ratio becomes 8.33 Watt power per N of weight force or 81.7 Watt per kg. Converted into funny units this is 0.05 HP per pound of mass. 1 HP can hold 20 lbs in the air, but that will not be enough to even spin up the propeller unless it can adjust blade pitch. For practical usability I recommend to double that number. If the propeller diameter shrinks and disk loading increases, efficiency will drop and the power required will rise.