Let $e_\lambda$ be an approximate unit of $I$. Say $e_\lambda$ is quasicentral if for every $a\in A$, $e_\lambda a-ae_\lambda\to 0$.
I have shown that when $A$ is separable, $I$ has sequential quasicentral approximate unit $\{e_n\}$ with $e_ne_{n+1}=e_n$. Here is the proof:
Let $e_\lambda$ be an arbitary approximate unit of $I$. for $\lambda$ and $\mathcal F\subseteq_{finite}A$ , define $H_\lambda$ to be the convex hull of $\{e_\eta:\eta\geq \lambda\}$ and define $S_{\lambda,\mathcal F}=\{\bigoplus_{a\in\mathcal F} (ha-ah):h\in H_\lambda\}$. Then the norm closure $\overline{S_{\lambda,\mathcal F}}$ contains $0$. Therefore there is an element $h\in H_\lambda$ such that $\left\|\bigoplus_{a\in\mathcal F}(ha-ah)\right\|<1/n$, where $n=|\mathcal F|$. Denote by $h_{\mathcal F,\lambda}$ such $h$.
Assume $h_{\mathcal F,\lambda}=\sum \alpha_ie_{\eta(i)}$ ,write $r(h_{\mathcal F,\lambda})=\max_i\eta(i)$, and define :
$$(\mathcal F,\lambda)\leq (\mathcal G,\eta)\Leftrightarrow \mathcal F\subseteq \mathcal G\text{ and } r(h_{\mathcal F,\lambda})\leq \eta $$
Then $h_{\mathcal F,\lambda}$ is a quasicentral approximate unit.
When $A$(or $I$) is separable, $e_\lambda$ can be chosen to satisfy that $\lambda\in\mathbb N$ and $e_ne_{n+1}=e_n$. Then $h_{\mathcal F,\lambda}h_{\mathcal G,\eta}=h_{\mathcal F,\lambda}$ when $(\mathcal F,\lambda)\leq (\mathcal G,\eta)$.Again since $A$ is seqarable, there are $\mathcal F(n),\lambda(n)$ such that $h_n=h_{\mathcal F(n),\lambda(n)}$ forms a sequential quasicentral approximate unit.
However, if $I$ is separable while $A$ is not, I don't know how I can modify the proof to get a sequential quasicentral approximate unit with $e_ne_{n+1}=e_n$.