I know the result in 1 variable, using the fact that localization of a PID is also PID. However, for $n>1$ variables, how can we prove it? (I think it is still true.)
The first thing I've got in my mind is the fact that, for given field $k$, $$k[x_{1},x_{1}^{-1}, \dots, x_{n},x_{n}^{-1}] \cong k[x_{1},y_{1}, \dots, x_{n},y_{n}]/ (x_{1}y_{1}-1, \dots, x_{n}y_{n}-1).$$ And it suffices to show that $(x_{1}y_{1}-1, \dots, x_{n}y_{n}-1)$ is prime. However, I have no idea how to show that this ideal is prime. Any hint will be appreciated.
It is not a PID (which is apparently what your question is asking) but it is a domain, (which is what your title is asking.)
$k[X,Y]$ has Krull dimension $2$, and if you localize at the multiplicative set generated by $X$ and $Y$, you're going at least to still have a tower of prime ideals $\{0\}\subseteq (X+1)\subseteq (Y+1)$, so the localization is still $2$ dimensional. But a PID must have Krull dimension $\leq 1$.