Fuses won't have the well defined cut-off at 3.001 A that you might be hoping for. It's more likely that the lamp will blow and protect the fuse. I don't think that any great complexity is worth the trouble to protect a relatively cheap lamp.
You can add parallel resistance to divert 1 A away from the bulb but this is a potential failure and a waste of energy (12 W).
Figure 1. The typical Mean Well PSU design. Source: Dimmable mains PSU control.
Instead I would be inclined to either wire a resistor securely in parallel with the dimmer pot to bring the maximum control voltage down to 7.5 V.
simulate this circuit – Schematic created using CircuitLab
Figure 2. The simplest current limiter for resistance control.
The constant current source, as best as I can make out, is a 0.1 mA so if the dimmer resistance control isn't allowed go above 75 kΩ you will have the current limit control required. (100k || 330k = 77k.)
Figure 3. The simplest current limiter for PWM control.
With Figure 3 we have limited the control voltage to 7.5 V max with R2. Applying PWM with Q1 will reduce the voltage which will be filtered by Figure 1's R1 / C1 low-pass filter. Note that with PWM at 0% the control voltage will be at the maximum (7.5 V = 75% of specified current) and with PWM at 100% the control voltage will be at minimum (0 V = 0% of specified current). i.e. You need to invert your PWM strategy.