How to initiate a ring oscillator
Not only can you do this, you can synchronize the start of oscillations quite precisely.

EDIT - Not only can you start the oscillator precisely, this topology will produce a complete last pulse. Simpler on-off configurations may produce narrow last pulses if the off condition occurs during a pulse, but not this one. END EDIT

Start with this structure







simulate this circuit &ndash; Schematic created using CircuitLab

A low input will inhibit oscillation, and when the input goes high oscillation will start.

As for the time delay, assuming you're using discrete logic, you can take advantage of the common occurrence of quad NAND gate ICs by implementing







simulate this circuit

Of course, you can also use a couple (or 4 or 6 or ...) of inverters to do the job.

Alternatively, you can simply brute-force the issue. Using an open-collector output or a separate transistor, you can







simulate this circuit

This takes into account that every (discrete, at least) logic family on the market will tolerate a short circuit to ground indefinitely. If it could not, you'd get any number of dead ICs due to inadvertant shorts during test, assembly or use. This approach will cause a slight change in the output frequency when oscillation first starts, due to the fact that the chip would probably get warm and this would affect the propagation delay of the gates. But this would quickly disappear as the IC reached operating temperature.