Yes, $I=Ap$ for some projection $p$. More generally, suppose that a closed left ideal
$I$ contains a sequence $c_1,c_2,\dots$ such that
$
\sum_{i=1}^\infty \|c_i\|<\infty
$
and for any $x\in I$ there exist $x_1,x_2,\dots$ such that $\|x_i\|\leq \|x\|$ and
$$
\sum_{i=1}^\infty x_ic_i=x.
$$
Then $I$ is generated by a projection. (In the question, $c_i=\|R_i\|b_i$ and
$x_i=R_i(x)/\|R_i\|$.)

Here is a proof: Let $c=\sum_{i=1}^\infty (c_i^*c_i)^{\frac 1 2}$ (which converges). Since $(c_i^*c_i)^{\frac 1 2}\leq c$, we have $c_i=d_ic$ for some contraction $d_i\in A^{**}$.
Choose $\epsilon>0$ small enough ($\epsilon=\frac 1 6$ will suffice). Choose $N$ such that
$\sum_{i>N}^\infty \|c_i\|<\epsilon$. Since $c^{1/k}\in I$ for all $k=1,2,\dots$, we must have
$$
c^{\frac 1 k} =\sum_{i=1}^N x_ic_i+\sum_{i>N} x_ic_i=Xc^{\frac 1 2}+\Delta,
$$
where $X=\sum_{i=1}^Nx_id_i$ and $\|\Delta\|\leq \sum_{i>N} \|x_i\|\|c_i\| \leq \epsilon \|c\|^{1/k}$. Then,
\begin{align*}
c^{\frac 2 k} &=c^{\frac 1 2}X^*Xc^{\frac 1 2}+ \Delta'\leq \|X\|^2c+\|\Delta'\|\cdot 1.
\end{align*}
The norms of $X$ and $\Delta'$ can be estimated: $\|X\|\leq \sum_{i=1}^N\|x_id_i\|\leq N\|c\|^{1/k}$ and $\|\Delta'\|\leq 5\epsilon \|c\|^{2/k}$ (probably not a sharp estimate; I'm skipping the details). So
$$
c^{\frac 2 k}\leq N^2\|c\|^{\frac 2 k}\cdot c+ 5\epsilon \|c\|^{\frac 2 k}\cdot 1.
$$
This inequality must hold in $C^*(c,1)$ for all $k$. For $k$ large enough, $\|c\|^{2/k}$ is close to 1, so the right hand side is bounded by a linear function in $c$ which at 0 is less than $6\epsilon$. So $0$ can only be an isolated point of the spectrum of $c$. Now,
with $p\in I$ the support projection of $c$, we have $c_ip=d_icp=d_ic=c_i$ for all $i$, and so $xp=x$ for all $x\in I$.