The easy case is when $L=0$. There are two possibilities for the domain of $y$: either $y\in [0,0]$ (if $x=0$) or $y\in [-U,U]$ (if $x=1$). One binary variable is enough to capture two choices, so we don't need any new variables. We just need to rewrite the constraint as$$-Ux\le y \le Ux.$$If $L>0$, then there are

*three*possibilities for the domain of $y$: $[-U, -L] \cup [0,0] \cup [L, U]$. That means we'll need at least two binary variables to cover all choices. Rather than change the interpretation of $x$ (which may be used elsewhere in the questioner's model), I'll introduce two new binary variables ($z_1$ for the left interval and $z_2$ for the right interval) and link them to $x$ via $x=z_1+z_2$. That leads to the following constraints:$$-Uz_1 \le y \le -Lz_1 + U(1-z_1)$$and$$Lz_2 - U(1-z_2) \le y \le Uz_2.$$ If $x=0$, both $z_1$ and $z_2$ must be 0, and the new constraints force $y=0$. If $x=1$, then either $z_1=1$ or $z_2=1$ (but not both). Left to the reader as an exercise: verify that $z_1=1\implies -U\le y \le -L$ while $z_2=1 \implies L\le y\le U$.