Let $A$ be a $C^*$-algebra. We can regard its double dual as the universal enveloping von Neumann algebra of $A$ as follows: $A$ sits inside of $A^{**}$ via $i: A \hookrightarrow A^{**}$, where $i(a)(\varphi)=\varphi(a)$ for any $\varphi \in A^*$. It's known that $i(A)$ is weakly-$*$ dense in $A^{**}$. Then, since the $C^*$-algebraic operations are continuous, they extend to $A^{**}$. These extensions turn $A^{**}$ into a Banach algebra; the $C^*$ identity also extends, making $A^{**}$ into a unital $C^*$-algebra.
Let $Z:=\{ \omega \in A^{**}: \omega i(a)=i(a)\omega \ \forall \ a \in A\}$ and $Z_u:=\{ \omega \in Z: \omega \text{ is unitary} \}$.
I just read something that seems to imply that the product of any element in $Z_u$ with an element of $i(A)$ is in $i(A)$. That is, $\omega i(a) =i(a)\omega \in i(A)$ for any $\omega \in Z_u$.
I am not entirely sure whether this is true and I can't come up with a simple argument to prove it. I will appreciate any help to figure this out.
It is not true in general.
Suppose $A$ is unital. Because $i(A)$ is dense, $Z$ is the centre of $A^{**}$. You would have $\omega=\omega i(1)\in i(A)$ for all $\omega\in Z_u$, and so for all $\omega\in Z$. So the centre of $A$ would be a von Neumann algebra.
In the abelian case, the condition implies that $i(A)$ is an ideal in $A^{**}$, which is not always the case. For example, take $A=c_0+\mathbb C\,1$. Then $A^{**}=\ell^\infty(\mathbb N)$. As everything is abelian, $Z=A^{**}$. If for instance we take $\omega=1-e_1$, then $\omega=\omega\,1\not\in A$.