Как да получа списък само с ObjectId, използващ pymongo?

Използвайте distinct

In [27]: agent_ids = agents_collection.find().distinct('_id')

In [28]: agent_ids
Out[28]: 
[ObjectId('553662940acf450bef638e6d'),
 ObjectId('553662940acf450bef638e6e'),
 ObjectId('553662940acf450bef638e6f')]

In [29]: agent_id2 = [str(id) for id in agents_collection.find().distinct('_id')]

In [30]: agent_id2
Out[30]: 
['553662940acf450bef638e6d',
 '553662940acf450bef638e6e',
 '553662940acf450bef638e6f']