In terms of an approach to this, I'd approach it in the following way, although there is a certain amount of guessing/estimation involved:

You should be able to find out the 'Max_propulsion_force' of the vessel, ie the maximum thrust developed by the propellors in Newtons. Either you can find this directly, or if you know the power, and the maximum speed, then the thrust must be (power[W]/speed[m/s]). Remember that 'Max_propulsion_force' is for a single engine.

When you know this, you also know that the total drag must equal the maximum propulsion force (from both engines if relevant) when the ship's at it's maximum speed.

If you want, it might be simplest to start assuming that all drag is proportional to the square of the speed (i.e. the DynamicsSpeedA term), and set the DynamicsSpeedB to zero. This allows you to say:

Total thrust [N] = DynamicsSpeedA * Maximum speed[m/s]^2, or

DynamicsSpeedA = Total thrust [N] / (Maximum speed[m/s]^2)

In reality, the total drag will be a combination of the 'A' and 'B' type, so you can adjust the A and B terms, so you still ensure that

Total thrust [N] = DynamicsSpeedA * Maximum speed[m/s]^2 + DynamicsSpeedB * Maximum speed[m/s]

Does this make sense? I have an excel sheet somewhere that might help, if that would be useful.

For the other terms, the same sort of approach applies. I would normally set the inertia from the known value (the polar moment of inertia about the ship's yaw axis), and then adjust the Rudder A and B terms to give a realistic start of turning motion (Rudder A sets the turning effect proportional to the ship speed, independent of the engine, so in an extreme example, would apply to a paddle steamer. Rudder B uses the engine speed, so accounts for the wash over the rudder from the propellor, which can turn the ship even if it is stationary). I'd then adjust the DynamicsTurnDrag terms to ensure that the steady rate of turn is reasonable for the rudder angle.

The lateral drag terms are used to account for side slip of the ship, and should be quite a bit higher than the normal drag terms.