I will assume familiarity with the statement, which can also be found here, and I will use the notation there too. http://en.wikipedia.org/wiki/Kaplansky_density_theorem
I have a problem with the standard proof, when people claim that it works at this level of generality where A need not have the unit $1_H$ of B(H). One usually proceeds first by reducing in the obvious way to where the $A$ is norm closed, and then to where we consider just the self adjoint elements Then they pick some continuous function from R to R with the property that it smoothly interpolates between the limits at $\infty$ and $-\infty$ of $+1$ and $-1$ respectively. For instance, one could use $f(t)=t/(t^2+1)$. In the midst of this argument, one has to use that the functional calculus takes the self adjoints in $A$ to the self adjoints in $A$, because the point of this function is basically to squash your strongly approximating net to be within the unit ball of the self adjoint subspace of A. This $f$ is a continuous function, so I may as well be talking about the continuous functional calculus as opposed to Borel, and then one observes that the continuous functional calculus relative to A is the same as relative to B(H) because the spectra coincide. That is how one sees that when I compute $f(a)$ relative to $B(H)$ for $a$ in $A$ self adjoint, I still end up in $A$. This all goes bad when $A$ doesn't contain $1_H$.
How can I understand why nevertheless, the statement is true at this level of generality?