This is false. Consider, for example, the case of M being the von Neumann algebra of bounded complex-valued functions on an infinite discrete set I. It acts on the Hilbert space of square-summable functions on I, which is the standard form of M. Take any ξ∈H that is everywhere nonvanishing (e.g., n↦1/n). Then any x∈M such that xξ=ξ must satisfy x=1.

This is false. Consider, for example, the case of M being the von Neumann algebra of bounded complex-valued functions on an infinite countable set I. It acts on the Hilbert space of square-summable functions on I, which is the standard form of M. Take any ξ∈H that is everywhere nonvanishing (e.g., n↦1/n). Then any x∈M such that xξ=ξ must satisfy x=1.

Much more generally, for any M that admits a faithful finite trace τ, the element ξ=τ^{1/2} is an element in the standard form of M and is a counterexample to your claim: if xξ=ξ, then (x−1)ξ=0, so rsupp(x−1)lsupp(ξ)=0, but lsupp(ξ)=lsupp(τ^{1/2})=1, hence rsupp(x−1)=0, i.e., x−1=0 and x=1.