The von Neumann algebra generated by a non-closable operator

The answer to Question 1. is positive. Namely, consider the polar decomposition of your operator $M=U|M|$ and define $X:= U f(|M|)$, where $f:[0,\infty) \to [0,\infty)$ is a bounded increasing function, for instance $f(x)= 1 - e^{-x}$ would do. Then the spectral projections of $f(|M|)$ are the same as the spectral projections of $|M|$, so both $U$ and the spectral projections belong to the von Neumann algebra generated by $X$, therefore it is the same as the von Neumann algebra generated by $M$.

Unfortunately, it does not give a clue about Question 2.

Only a comment on Question 2: It is already not easy to define the von Neumann algebra generated by a family of closable operators, see e.g. https://projecteuclid.org/euclid.cmp/1103899047 Def. 2.5 and Rem. 2.7.

Closability in this context appears naturally for the following reason: Assume that you have some (possibly non-closed) unbounded operator $M$ on some dense domain $\mathcal D$. To get towards defining a corresponding von Neumann algebra $\mathcal M$, I presume you would first try to compute a formal adjoint $M^*$ and I think you will agree that $M^*$ should also be densely defined. But then you already obtain that $M$ is closable.