This is a refinement of Iosif Pinelis's answer, so we shall be somewhat brief. For a punchline, jump to the "Added" section below.
We claim that if $a_n>c/n$ holds some $c>1$ and for all $n\geq n_0$, then $P(I=S^1)=1$. To see this, fix a large integer $N$ and any interval $J$ of length $1/N$. Then, $$P(J\not\subseteq I)\leq\prod_{n_0\leq n<cN}\left(1-\left(\frac{c}{n}-\frac{1}{N}\right)\right)<\exp\left(\sum_{n_0\leq n<cN}\left(\frac{1}{N}-\frac{c}{n}\right)\right)=O_c(N^{-c}).$$ As $S^1$ is a union of $N$ intervals of length $1/N$, $$P(I\neq S^1)=O_c(N^{1-c}).$$ The right hand side tends to zero as $N\to\infty$, hence $P(I\neq S^1)=0$ as claimed.
Added. Using Google I found out that Shepp (1971) aswered the question completely (see here). In particular, $P(I=S^1)=1$ holds when $a_n\geq 1/n$ for all $n\geq n_0$.