$\newcommand{\ep}{\varepsilon}$

Without loss of generality (wlog), $a_1\ge a_2\ge\cdots$ and $a_n\to0$ as $n\to\infty$. 
Then a sufficient condition for $P(I=S^1)=1$ is that for some real $k>3/4$ and all large enough $n$ we have $a_n\ge\frac kn\,\ln n$. Indeed, take any real $\ep\in(0,1/2)$ and let  
\begin{equation}
	n_\ep:=\max\{n\colon a_n>3\ep/2\},  
\end{equation}
so that 
\begin{equation}
	n_\ep\ge\frac{2k-o(1)}{3\ep}\,\ln\frac1\ep; 
\end{equation}
all the limits here are taken for $\ep\downarrow0$. 

Take any interval $J$ of length $\ep$, centered at some point $x$. Then 
\begin{equation}
	P(J\not\subseteq I)\le P(d(X_j,x)\ge\ep\ \forall j\le n_\ep)
	=(1-2\ep)^{n_\ep},
\end{equation}
where $X_j$ is the center of the random interval $I_j$ and $d(X_j,x)$ is the distance from $X_j$ to $x$. Let now $J_1,\dots,J_{m_\ep}$ be intervals of length $\ep$ each such that $J_1\cup\dots\cup J_{m_\ep}=S^1$; wlog, $m_\ep\sim1/\ep$. 
Thus,
\begin{equation}
	P(I\ne S_1)\le\sum_{i=1}^{m_\ep} P(J_i\not\subseteq I)\le m_\ep(1-2\ep)^{n_\ep}
	\le\frac{1+o(1)}\ep\,\exp\Big\{-\frac{4k}{3+o(1)}\,\ln\frac1\ep\Big\}\to0,
\end{equation}
so that $P(I\ne S_1)=0$, as claimed.