Why can we assume the Von Neumann Algebra contains the identity? Consider the following fragment from "Murphy's $C^*$-algebras and operator theory":

The proof of theorem 4.1.11 says we can assume $id_H \in A$ in (1). I assume this to be given and from this I want to deduce that the result also holds when $id_H \notin A$.
If $id_H \notin A$, we have $$A \cong A_p$$ by remark 4.1.2 where $p$ is the unit of $A$. Moreover, the von Neumann algebra $A_p\subseteq B(p(H))$ contains $id_{p(H)}$ so by what is already known we know that $A_p$ is the closed linear span of its projections. Since we have a $*$-isomorphism $A \cong A_p$, the same also holds for $A$.
Is the above correct? I'm self-studying this and some feedback from time to time really helps me.
 A: What you are doing is correct. I'll briefly repeat your construction. Suppose $A$ is a von Neumann algebra with underlying Hilbert space $H$, it is standard that $A$ admits a unit, call the unit $p$. Note that $p$ is necessarily an orthogonal projection in $B(H)$.
For any $a\in B(H)$ define $a_p := pap$, which we understand as a map $p(H)\to p(H)$. This gives a linear map $B(H)\to B(p(H))$ and if you restrict this map to $A$ it becomes an injective $C^*$-morphism (since $p$ is the unit of $A$). It follows that the map is a $C^*$-isomorphism from $A$ to the image of $A$, call this image $A_p$.
Note that $A_p$ is a von Neumann algebra with underlying Hilbert space $p(H)$, I think the easiest method would be to check that it is a SOT closed subspace of $B(p(H))$. Further $p_p=\mathrm{id}_{p(H)}$, so $A_p$ contains the identity of the Hilbert space onto which it is represented.

Now if you would like to check some property of $A$ that does not explicitly reference the representation on $H$ (ie it is a property of $C^*$-algebras that are isomorphic to a von Neumann algebra rather than concrete von Neumann algebras) you can work with $A_p$ instead of $A$. For example if you can show that a von Neumann algebra containing the identity is generated by projections then this holds for $A_p$, but $A_p$ is isomorphic to $A$ and the property "is generated by projections" will be carried over by the isomorphism.
