Solution:
$(1)$ Take $M=1$. For any $x \in \mathbb{R}$, $g(x+M)=\sin (\pi x+\pi)=-\sin \pi x=-g(x)=Mf(x)$. Therefore, $g(x) \in P$.
$(2)$ When $M=1$, $f(x+1)=-f(x)$ and $f(x+2)=-f(x+1)=f(x)$. Therefore, $f(x)$ is a periodic function with a period of $2$.
$(3)$ Since $h(x)=\sin \omega x \in P$, there exists a non-zero constant $M$, such that for any $x \in \mathbb{R}$, $h(x+M)=-Mh(x)$ holds, i.e., $\sin (\omega x+\omega M)=-M\sin \omega x$.
If $|M| > 1$, taking $\sin \omega x=1$, then $\sin (\omega x+\omega M)=-M$ cannot always hold for $x \in \mathbb{R}$.
If $|M| < 1$, taking $\sin (\omega x+\omega M)=1$, then $\sin \omega x=-\frac{1}{M}$ also cannot hold for $x \in \mathbb{R}$. Therefore, $M=\pm1$.
When $M=1$, $\sin (\omega x+\omega)=-\sin \omega x$, $\sin (\omega x+\omega)+\sin \omega x=0$, $2\sin (\omega x+ \frac{\omega}{2})\cdot \cos \frac{\omega}{2}=0(x \in \mathbb{R})$, $\cos \frac{\omega}{2}=0$ yields $\omega=2k\pi+\pi(k \in \mathbb{Z})$;
When $M=-1$, $\sin (\omega x-\omega)=\sin \omega x$, $\sin (\omega x-\omega)-\sin \omega x=0$, $2\cos (\omega x- \frac{\omega}{2})\cdot \sin (- \frac{\omega}{2})=0(x \in \mathbb{R})$, $\sin \frac{\omega}{2}=0$ yields $\omega=2k\pi(k \in \mathbb{Z})$.
In conclusion, the range of $\omega$ is $\boxed{\omega=k\pi(k \in \mathbb{Z})}$.