How would I go about calculating the circuit load with my x10 probe?
Loading is relative to the system impedance, which is determined by the pin output resistance (at low frequencies), or the transmission line (at high frequencies). Simple resistor divider rules apply, give or take.
Note that a low-frequency approximation would treat the cable as a capacitance, but the impedance does not approach zero as frequency goes up. That's the magic of transmission lines!
The same is true of the probe, which ultimately is a transmission line as well. It uses a lossy TL of particular impedance and length, so that high frequencies are attenuated by about 10x when driven with what's in the probe tip. (When in 1x mode (if present), the tip compensation network is bypassed, line losses dominate, and the cutoff frequency is around 10MHz!) The impedance, loss and equivalent capacitance determine what RLC network is needed for 10x compensation -- no, it's not simply a 9M resistor! It's more like 9M in parallel with (~10pF + 100 ohm) or something like that: the resistance increases HF loss if needed, and the cap allows the probe's impedance to smoothly transition from 9M to ~100s ohms over a long frequency range (~3kHz to 30MHz) while keeping a square step response (as measured from a voltage generator!).
So, it follows that the probe looks like ~100s ohms in the "high frequency" range. This will have some impact on the cable -- but, with only a 100MHz or so scope, you wouldn't be able to tell the difference. (If you were getting bit errors, it might be enough damping to prevent errors, thus illustrating a case of "observer effect"...)
It's also a good illustration that a "passive Zo" probe can do a better job: a 450+50 ohm resistor divider (450 ohms at the probe tip; the cable serves as the lower resistor, and can be any length of ideal 50 ohm cable, terminated at the scope) has a relatively low impedance at DC (several orders of magnitude worse than the usual probe!), but comes out significantly better at high frequencies. 500 ohms out of a 100 ohm system is only a 17% reduction in impedance (and thus a similar amount of damping, reflection, SWR, however you want to account for it). It's more than 10% (the general threshold of "not caring"

), but not terrifically more: so, mind the difference and try to anticipate its effects, but what you see won't be terrifically far off from reality.
This may seem a stupid question but is it possible for me to include the resistor on the receiver side and not the transmitter side?
Yes, but it has to be a parallel resistor (to ground), which is obviously a bit of a problem!
You can use an R+C where R is characteristic and C is >= 2.5 x C_line (which includes cable, trace and pin capacitances). But this is impractical for long lines and high clock rates, because it has a lowpass filtering effect and dissipates a lot of power (which is hard on the resistor, and even harder on the driver!).
It's probably pretty reasonable here, as you'd only need maybe 47pF?
If the clock rate is much slower than the RC time constant, then it will be fully charged/discharged every cycle, and the total ESR (includes pin driver and termination resistor) will dissipate P = (0.5 * C * VCC^2) * 2 * Fclk.
Tim