The following is a theorem of Murphy's C*-algebra and operator theory:
I think it can prove easier, while I'm not sure about my proof :
Let $a\in A$ be positive. Consider $C^*(a)$, and let $\{u_i\}$ be its approximate unit.
We know $u_i\to p$ (sot) while $p$ is the unit of $von(a)$. Because $pa=ap$, we have $\overline {aH}\subset pH$.
Also $u_i$ is generated by $a$, so $\overline{u_iH} \subset \overline{aH} $, and therefore $pH=\cup\overline{u_iH} \subset \overline{aH} $, which shows that $p=[a]$.
Please help me. Thanks in advance.
I think your argument is fine, with a caveat on the word "easier". Your assertion "we know $u_i\to p$ (sot)" is basically what Murphy's proof is about.
On a minor detail, the assertion $pH=\bigcup \overline{u_i H}$ requires that $u_i\nearrow p$, which you do have in this case. Otherwise you have inclusion, which still works in your case.