How general is the following statement about Murray-von Neumann equivalence of projections in a von Neumann algebra?
Let $M$ be a von Neumann algebra and let $p,q\in M$ be projections. If there exists a sequence $\{x_n\}_{n\in\mathbb N}\subset M$ such that $$p=\operatorname*{{\scriptsize SOT}\ lim}_{n\to\infty}x_n^*x_n\qquad\text{and}\qquad q=\operatorname*{{\scriptsize SOT}\ lim}_{n\to\infty}x_nx_n^*$$ then $p$ and $q$ are Murray-von Neumann equivalent projections.
This seems to be true at least when $M$ is a factor and $\tau$ is a trace on $M$, for in this case $p\sim q$ iff $\tau(p)=\tau(q)$.
It fails badly on $B(H)$: let $p=I$, $q=0$, and $x_n=\sum_{k=1}^\infty E_{n+k,k}$, where $\{E_{k,j}\}$ is a set of matrix units. Then $$ x_n^*x_n=\sum_{k,j=1}^\infty E_{k,n+k}E_{n+j,j}=\sum_{k=1}^\infty E_{kk}=I, $$ $$ x_nx_n^*=\sum_{k,j=1}^\infty E_{n+k,k}E_{j,n+j}=\sum_{k=1}^\infty E_{n+k,n+k}=\sum_{k=n+1}^\infty E_{kk}\xrightarrow[n\to\infty]{sot}0. $$ This makes it fail in any infinite von Neumann algebra.
As you mention, it works on finite factors. I'm not sure about finite von Neumann algebras, because there equal trace is not enough for equivalence.