Is this DO LET sugar considered pure Lambda calculus and nullipotent / side-effect-free?
Anything that can be encoded in the pure lambda calculus, as you realize in the last paragraph, can not have side effects. Otherwise, the lambda calculus itself would be able to express those side effects as well.

However, note that your encoding looks a bit weird. For instance,

(LET a 1 b 2 e) = (λ a . (λ b . e) 1 2)


does not bind 1 to a and 2 to b. You probably want instead

(LET a 1 b 2 e) = (((λ a . (λ b . e)) 1) 2)


which does the correct binding. The other encoding should be fixed similarly.

I would suggest you "test" these encoding by performing $\beta$ steps. If you don't reach $e\{1/a,2/b\}$ something looks wrong.