Your second if-then statement is always true because $Y$ is binary. For your first if-then statement, rewrite as its contrapositive $Y=0 \implies tS \ge \epsilon$. The following big-M constraint enforces that:
$$\epsilon - tS \le MY$$
This is equivalent to what you tried. Note that $(tS,Y)=(\epsilon,0)$ is feasible, so if the solver always returns it, maybe it is optimal. As a sanity check, you could fix $tS$ to $0$ and see what happens.
Update based on a lower bound of $0$ for $tS$: you can take $M=\epsilon-0=\epsilon$, yielding $$tS\ge\epsilon(1-Y)$$