As I was trying to figure out this question, I wondered if the following is true:
If $A$ is a non-unital $C^*$-algebra and $\tilde{A}$ denotes its unitization, then is it true that any $*$-subalgebra $B\subset\tilde{A}$ that is dense in $\tilde{A}$ satisfies $B\cap A\neq\{0\}$?
It seems to me that this is too much to ask and I cannot see any particular reason for this to be true. Of course, if $1_{\tilde{A}}\in B$, then the result is obvious. Also, (maybe this helps), by approximating the unit through $B$ we get a sequence of invertible elements that belong to $B$, but we cannot deduce anything about their inverses (whether they belong to $B$ or not). Note that this argument proves the result if we also assume that $B$ is a (one-sided, in general) ideal in $\tilde{A}$.
On the other hand, I cannot think of any counter-example.
I am particularly interested for the abelian case, because non-unital abelian $C^*$-algebras are continuous functions over LCH spaces, so we can identify the unitization with the continuous functions over the one point compactification and maybe some topological arguments can prove/ disprove the result.
By definition, $A$ is an ideal in $\tilde A$. Consider the quotient map $\pi\colon \tilde A\to \tilde A/A\cong \mathbb{C}$. By definition, $x\in A$ if and only if $\pi(x)=0$. If $A\cap B$ were trivial, then $\pi$ would be injective on $B$, which clearly cannot be the case unless $\dim B\leq 1$. And it's not hard to see that $\dim B\leq 1$ can only occur under these circumstances if $A=\{0\}$, $\tilde A=\mathbb{C}$ and $B=\tilde A$. Otherwise $A\cap B\neq\{0\}$.