Let me try to do this systematically. The data sheet is somewhat confusing.
I'm reading all three transistors to be switches. If switch 1 is on, 2
will be on and 3 off. Thus Vout=Vcc. If switch 1 is off, 2 will be off
and 3 on, Vout=GND.
Applying 3.3v and 1mA source causes a voltage drop of around 2V from
my DC power supply.
Looking at the circuit, I think that max output is always limited to
Close enough, but not quite. If Q1 is off, Q3 will be off, and Q2 will be on - sort of. Q2 will get its base drive from Vcc, but the transistor cannot turn on really hard. So you should figure on an output in the range of Vcc minus 1 to 2 volts, depending on the current being supplied. And this, I think, is consistent with your providing 3.3 Vcc and getting 2 volts out, although your description is not clear.
Note that the data sheet says that the 550 "will drive up to 8 TTL loads". TTL input levels are not the same as CMOS, and a low TTL input is 0.8 v, while a high input is 1.6 volts. So a driver intended for TTL does not worry a lot about the high output voltage.
It says that the maximum input voltage is 10V, but the maximum voltage
output is 35V.
It sure does, but you need to learn how to read data sheets. These numbers come from page 4 of the data sheet, and this is labeled "Absolute Maximum Ratings". The numbers on this page should not (must not) be used for operation. They simply give limits that the manufacturer says the chip will survive. So, for instance, if you apply a Vcc of 10 volts to your 550, all this says is that you won't cause the chip to explode, and if you then turn the voltage down to something more reasonable, like the 4.5 to 5.5 listed on page 5, the chip should still work.
Likewise, the 35 volt output limit has a note attached - see the little (4)? Looking down at the bottom you'll see that (4) says that 35 volts can only be applied directly to open-collector types, and the 550 is not one of those. If the output might somehow be attached to a higher voltage, you need to put a resistor between the two to limit the current that the output has to handle.
The upshot of this is that you should provide 5 volts to the 550. At 3.3 it might work, and it might not, and it might sorta, kinda work. Page 5 says that the 550 should be operated in the range of 4.5 to 5.5 volts, and if you run it with ripple of less than 2 volts (that is, within the range of 4 to 6 volts) it won't false-trigger on the ripple. More than that, and it still might not false-trigger, but it might, and you've been warned. Running outside the 4.5 to 5.5 range is not guaranteed to cause problems, but the chip is not guaranteed to work to the specifications provided, either. You can do it, and it will probably work in the sense of detecting some light level shifts, but if it's slower than you expect, or less sensitive, you have no ground for complaints, so don't go whining to the manufacturer or bad-mouthing the part on internet forums.
Finally, I hope you're aware that this thing is grossly oversensitive for what you want. From page 5, the Positive-Going Threshold Irradiance (light trigger level, in plain English) is in the range of 0.25 to 1.4 mW/$cm^2$. Multiplying by 10,000 gives 2.5 to 14 watts per square meter, about 1000 times less than you've specified. This may or may not cause false triggers (I don't know enough about your system), but you might consider putting an optical attenuator on the sensor to reduce sensitivity.