A possibly unbounded operator $T$ on a Hilbert space $\mathcal H$ is (in my source) defined as affiliated to a von Neumann algebra $M$ if for each unitary element $u$ of $M^\prime$, $u^*Tu=T$ (or moreover, $u$ commutes with $T$). For $T \in B(\mathcal H)$, this should imply that $T \in M$, or by the bicommutant theorem, that $T$ commutes with all $x \in M^\prime$ and not just its unitary elements.
This seemed easy, but I am not seeing how this is true right now. I've also messed around with trying to prove that $T$ is in the WOT-closure of $M$, again without success. Can anyone help me out here?
Every element $x$ of a C*-algebra is a linear combination of unitary elements of the algebra. To see this, note that $x = \frac12(x+x^*)+i\left(\frac{1}{2i}(x-x^*)\right)=\mathrm{Re}(x)+i\,\mathrm{Im}(x)$ is a linear combination of self-adjoint elements, so it is enough to show it for the case where $x=x^*$. We may rescale and assume that $\|x\|\leq 1$. Then $x=\frac12(x+i\sqrt{1-x^2})+\frac12(x-i\sqrt{1-x^2})$ is the average of the unitary elements $x\pm i\sqrt{1-x^2}$.
Since $M'$ is a C*-algebra, if an operator commutes with every unitary element of $M'$, then it commutes with every element of $M'$.