This is embarrassing, I think it must work, but I can't see how to prove it works. If anyone knows enough functional calculus of operators on a Hilbert space to tell me how to do it, I would be very grateful. Of course, you might also tell me that it is wrong...

Take positive bounded operators $x,y$, and increasing continuous functions $f,g:[0,\infty)\to [0,1]$ with the property that if $f(t)>0$ then $g(t)=1$. Then show that $$ f(x)\,g(x+y)= g(x+y)\,f(x)=f(x)\ . $$

Reason why I think it should work: If $f(t)>0$ then $g(t+\mathrm{positive})=1$, so it works... OK, this argument is not legal, but can the statement be proved in functional calculus?

Purpose: This is to be used to show that for a separable C* algebra there is an approximate identity $u_n$ with the property that $u_n u_{n+1}=u_{n+1}\ u_n=u_n$. Maybe someone actually looked at special approximate identities with this sort of property?

OK, the conjecture is false, as noted below. However there was a comment below about the property of approximate identities and whether this might still be true. After some thought I think that the problem is related to functional calculus of several commuting operators. Or rather to this modification: What happens to the functional calculus of several bounded operators which only almost commute, $f(T,S)$ where $\|[T,S]\|<\epsilon$? For an approximate identity we do have for fixed $n$, $\|[u_n,u_m]\|\to 0$ as $m\to\infty$, so it might still be possible to formulate the construction if a small commutator does not matter...

Yes, making this into a general note about functional calculus of two non-commuting operators is a but of an extrapolation.....

NOTE - I have just read the comment on the existence of these approximate identities below - this certainly solves my immediate problem. I now have to find a reference. Many thanks to all involved! I am still curious about the two non-commuting operators though.