What you propose is the simplest star topology in existence.
RS-485 specifically advises against going for a star topology (mostly due to basically making it ambiguous where termination should be, and there's no "good" answer to that question); but at three full devices on the bus, things might be OK. What you need to realize is that there's 100m of cable between Device 1 and Device 2.
Your devices will have to be able to drive 100m of cable, and you'll need to find the sweet spot between having too little load at each node, and too much load overall in termination to avoid problematic echoes. You're building a frequency-selective channel here. Assuming the E-field change travels at \$\frac23 c_0=2\cdot10^8\frac{\text{m}}{\text{s}}\$, it needs 0.5 µs to travel the from Device 1 to Device 2 (100m), and 0.75 µs to travel back to the Master in the middle if it's reflected at the end. Now, if these 0.75 µs happen to be an odd multiple of half a signal period, you'll have destructive interference there. Uh oh. That means you'll lose signal integrity for some signalling frequencies (including subharmonics introduced through repeating bit patterns), and get amplifications for others. That's why termination is important!
Also: If you have some "collision avoidance" scheme where access to the shared bus can only be asserted when there's been silence for a specific amount of time, then you'll need to account for the travel time when calculating the minimum amount of wait-before-send.