Will it be still distinct?

The second and fourth equations will always be 0 and 1 and the second and fourth equations will be greater than 1 for $n>1$ . So the only condition we need is $n+n \neq n*n \implies 2n \neq n^2 \implies n \neq 2$ . So for any $n> 3$ , four distinct integers will be produced on the right-hand​ side.

For $n \in \mathbb N$ , we note that $n-n=0$ and $n \div n = 1$ for all $n$ . The two numbers are always distinct for all $n$ . For $n+n = n \times n = n-n$ , $\implies n=0 < 3$ . We note that $n+n \ge 2 \ne n \div n$ and that for $n \times n = n \div n$ , $\implies n = 1 < 3$ . And the last case is $n+n=n\times n$ , $\implies n=2<3$ .

Yes, always for integer greater than 3, the four equations will produce four distinct numbers.