Does a constant current source add to existing current

You mention that you want fast switching. None of the devices will be able to do that. Any the feedback circuits internally are designed to be slow so that they don't to much incorporate noise. And in fact some of the feedback is thermal feedback. Put another way, the dominant pole is low, 100's of KHz is typically the the fastest response time. This is true whether you are using a disable pin or connecting/opening the load to the source.

The best way to fix this is that you build an external circuit that current steers between two limbs. On one limb will be your LED chain and on the other a dummy load that draws the same amount of current (and hopefully also has similar load characteristics). An example of such a circuit is a differential pair. You should be able to get 10's of ns switching time then.

The challenge with the two limbs will be that the load characteristic differences may disturb the feedback of the LM317 etc. which then reacts slowly. If you can't match the two limbs then I'd suggest building a current mirror that decouples the current output the LM317 etc. so that it only sees a constant load the mirror transistor sees the variability.

You should be able to simulate all of this in LTSPice or similar before building.

The other factors you mention above not show stoppers, so pick what you want.

Recall that, for node voltage analysis, a floating voltage source (a voltage source that does not connect to the GND node) poses a problem since you cannot write an equation relating the current through to the voltage across.

What you must do then is enclose the floating voltage source in a supernode, which reduces the number of KCL equations by one, and add the equation relating the voltage difference between the nodes the voltage source is connected to.

Now, the dual of node voltage analysis is mesh current analysis and here we have the dual problem when we have a current source common to two meshes - we can't write an equation relating the current through to the voltage across a current source.

What must be done then is to form a supermesh which reduces the number of KVL equations by one and add the equation relating the difference of the mesh currents to the common current source.

So, write KVL counter-clockwise around the supermesh consisting of the two voltage sources and the two resistors

$$V_1 = I_aR_1 + V_2 + (I_b - I_c)R_2$$

You have, by inspection (no KVL required for this mesh - this is dual to no KCL required for the node connected to a non-floating voltage source)

$$I_c = -1.25A $$

You need one more equation which is the equation relating to difference of the two mesh currents with the common current source.

$$3A = I_a - I_b $$

Now, you have 3 independent equations and 3 unknowns.