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Feb 9, 2021 at 6:14 comment added admissiblecycle Thanks! This is a nice argument. After posting this question, I have learned that one can also deduce this statement from strong approximation for $S^3$. Proposition 2.4 in this survey -- birs.ca/workshops/2013/13w5019/files/… -- shows that strong approximation holds for $S^3$ away from any prime $p \neq 2.$ Then it follows that $S^3(\mathbf{Z}[\tfrac{1}{p}])$ is dense in $S^3$. But as your answer shows, this is perhaps a bit of an overkill.
Feb 9, 2021 at 6:04 vote accept admissiblecycle
Feb 8, 2021 at 17:43 history edited R.P. CC BY-SA 4.0
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Feb 8, 2021 at 16:38 history answered R.P. CC BY-SA 4.0