I'm reading the book "An invitation to $C^*$-algebra's" by William Arveson and I'm focused on the proof of proposition 1.3.1, which says:
Let $A$ be a $C^*$-algebra and $J$ be a closed two-sided ideal of $A$. Then for every $x \in J$ there is a sequence $(e_n)_n$ of self-adjoint elements of $J$ with
$$Sp_A(e_n) \subseteq [0,1], \quad \lim_n \Vert xe_n - x \Vert = 0$$
The author provides a proof in the case that $A$ has a unit and concludes by saying that the case where $A$ has no unit follows by adjoining a unit, leaving the details to the reader.
So, suppose $A$ has no unit. Then consider the unitalisation $A_I:=A \oplus \mathbb{C}$.
Maybe I can show that $J$ is a two-sided ideal in $A_I$?
So, let $a + \lambda 1 \in A_I$ and $x \in J$ ( $a \in A, \lambda \in \mathbb{C}$). Then
$$(a+\lambda1)x = ax + \lambda x$$
and if I can show that $\lambda x \in J$, I can show that $J$ is also a two-sided ideal in $A_I$.
However, I don't see why this should be true.
Note that $A$ is a closed two-sided ideal in $A_I$, and that $A_I$ is unital. So there is an approximate $e_n$ unit in $A$ by what you have already seen. Note that $e_n x\to x$ for all $x\in A$, from which follows that $xe_n=(e_n x^* )^* \to (x^*)^*=x$ by continuity of the $*$-operation (ie a left approximate unit is also a right approximate unit).
Now let $J$ be a closed left ideal (in the ring-theoretic sense) of $A$ (or right ideal, or two-sided). It follows that $\lambda e_n x\in J$ (resp. $x (\lambda e_n)\in J$) for all $x\in J$. Since $J$ is also closed you have that $\lambda x= \lim_n \lambda e_n x\in J$ (similarly $\lambda x = \lim_n x(\lambda e_n)\in J$).