Problem: Let $e_1, e_2, f_1$ and $f_2$ be projections of $M$ which is a Von Neumann algebra, such that $e_1e_2=f_1f_2=0$. If $e_1+e_2=f_1+f_2$, $e_1 \sim e_2$ and $f_1 \sim f_2$, prove that $e_1 \sim f_1$.
My attempt: Suppose $e_1$ and $f_1$ are not equivalent then by the comparison lemma there is a central projection $p$ such that either $pe_1\prec pf_1$ or $pf_1\prec pe_1$ (In this notation, for example $pe_1\prec pf_1$ means $pe_1$ is equivalent to a subprojection of $pf_1$ but $pe_1$ and $pf_1$ are not equivalent). Let's assume $pe_1\prec pf_1$. Then we have: $$pe_2 \sim pe_1\prec pf_1\sim pf_2.$$ Thus $pe_2\prec pf_2$. Moreover, since $p$ is central and $e_1e_2=f_1f_2=0$, we infer that $pe_1+pe_2\preceq pf_1+pf_2$. I wonder if we are allowed to replace $\preceq$ with $\prec$ in the last relation. At this stage, I don't know how to arrive at a contradiction while we already have $pe_1+pe_2=pf_1+pf_2$.
Any help would be highly appreciated.
It looks to me like this approach cannot work. To replace $\preceq$ with $\prec$ what you need is precisely to solve the original question! Since you want to say that $pe_1+pe_2=pf_1+pf_2$ contradicts $pe_1\prec pf_2$.
Now here is what comes to mind.
We have $e_1\sim 1-e_1$, $f_1\sim 1-f_1$. Since $e_1+e_2=f_1+f_2$, for simplicity we may assume that $e_1+e_2=f_1+f_2=1$, as everything happens below this projection (i.e., we can work on the von Neumann algebra $(e_1+e_2)M(e_1+e_2)$).
There exists a central projection $q$ such that $qe_1$ is properly infinite, and $(1-q)e_1$ is finite. This allows us to consider separately the cases where $e_1$ is finite, and where $e_1$ is properly infinite.
If $e_1$ is finite, so is $1-e_1\sim e_1$, and then $1=e_1\lor(1-e_1)$ is also finite. Let $p$ be a central projection with $pe_1\preceq pf_1$. We immediately get $p(1-e_1)\preceq p(1-f_1)$. So there exist projections $g_1,g_2$ with $g_1\leq f_1$, $g_2\leq 1-f_1$, and $pe_1\sim pg_1$, $p(1-e_1)\sim p g_2$. Then $$ p=pe_1+p(1-e_1)\sim pg_1+pg_2\leq pf_1+p(1-f_1)=p. $$ As $p$ is finite, this implies $pg_1+pg_2=p$. So $p(f_1-g_1)+p(f_2-g_2)=0$; this is a sum of projections, so $pg_1=pf_1$, $pg_2=pf_2$. It follows that $pe_1\sim pf_1$. In other words, $pe_1\prec pf_1$ is impossible. As can also also reverse roles, we get that $e_1\sim f_1$. So we have proven that $(1-q)e_1\sim (1-q)f_1$.
It remains to deal with $qe_1$, which is properly infinite. We have that $q$ is properly infinite; indeed, if $p\leq q$ is central and finite, then $pqe_1$ is finite so $pqe_1=0$ since $qe_1$ is properly. By the equivalence we have $pq(1-e_1)=0$, and so $p=pqe_1+pq(1-e_1)=0$.
Note that $qf_1$ is also properly infinite; if it weren't there would be a central projection $p\leq q$ with $pqf_1$ finite. By the equivalence we would have $pq(1-f_1)$ finite and then $p=pqf_1+pq(1-f_1)=p$ finite. This in turn makes $pqe_1$ finite, and then $p=0$ as above.
By this answer that you probably know about, we have $$ qe_1\sim qe_1\lor q (1-e_1)=q=qf_1\lor q(1-f_1)\sim qf_1. $$
Finally, from $qe_1\sim qf_1$ and $(1-q)e_1\sim(1-q)f_1$ we get $e_1\sim f_1$.