The answer is yes, even if $X$ is not normal. It even works for any connected solvable group acting on an affine variety with an open orbit.

To see this let $Y_1,\ldots,Y_r$ be the irreducible components of $X\setminus X^0$. Since the connected solvable group $B$ acts rationally on the ideal $\mathcal I(Y_i)\subset\mathcal O(X)$ it contains a non-zero $B$-semiinvariant $f_i$. Put $f:=\prod_if_i$ which is also a non-zero semiinvariant. Then $f$ vanishes on $X\setminus X^0$ by construction. On the other hand, if $f(x)=0$ with $x\in X^0$ then $f(X^0)=f(Bx)=\chi_f(B)f(x)=0$ which is impossible since $f\ne0$ and $X^0$ is dense in $X$. Hence $X^0$ is precisely the non-vanishing set of $X$.