m_ISS ~ 4.5e5 kg [1]
Rocket equation [2]:
m_0 = m_f exp(v_delta / v_e)
where
m_f = final mass, i.e. mass of ISS and the boosters
m_0 = m_f + propellant mass
v_delta = velocity change
v_e = effective exhaust velocity of the boosters
Let's try a high-thrust transfer from LEO to the Lunar Gateway's orbit via TLI (Trans-Lunar Injection) [3]:
v_delta = 3.20 + 0.43 = 3.63 km/s
For boosters, let's use the dual-engine Centaur III (because Wikipedia has mass and v_e data for it) [4]:
m_dry = 2462 kg
m_propellant = 20830 kg
v_e = 4.418 km/s
The idea is to attach n of these to the ISS. The rocket equation becomes
m_ISS + n (m_dry + m_propellant) = (m_ISS + n m_dry) exp(v_delta / v_e)
Solve for n:
n = m_ISS (exp(v_delta / v_e) - 1) / (m_propellant + m_dry (1 - exp(v_delta / v_e)) )
Plug in numbers and find
n ~ 32.4
So we need 33 Centaur III (and some way to attach them, which I optimistically assume won't add significantly to the ISS mass).
Total Centaur III + propellant mass: 33 * (2462 + 20830) = 768636 kg
Planned Starship payload capacity to LEO is 2e5 kg [5], so assuming that a way can be found to fit 7 Centaur III in its payload bay, we can get all 33 boosters to LEO with five Starship launches.
Why not use Starship itself? Its Raptor Vacuum engines have lower v_e (~3.7 km/s) [6], and if you want it back, you need to add fuel for the return trip to m_f. Exercise for the reader!
[1] https://en.wikipedia.org/wiki/International_Space_Station
[2] https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
[3] https://en.wikipedia.org/wiki/Delta-v_budget#Earth_Lunar_Gat...
[4] https://en.wikipedia.org/wiki/Centaur_(rocket_stage)
[5] https://en.wikipedia.org/wiki/SpaceX_Starship_(spacecraft)
[6] https://en.wikipedia.org/wiki/SpaceX_Raptor#Raptor_Vacuum