I think this is not an answer but a precisation. Assume that the statement is true for some $n$.
~~Then S^{n-1}\times SO(n-1)has a Lie group strcture and in particular a parallelizable tangent bundle. This implies that the tangent bundle of S^{n-1} is parallelizableand this is true if and only if n=2,4,8 (the original reference is //projecteuclid.org/euclid.bams/1183522319).~~ Edit: As explained below (by Bertram Arnold), this implies that the sequence

$SO(n-1)\to SO(n)\to S^{n-1}$

splits, and this split give rise to a section in the frame bundle of $S^{n-1}$, i.e it gives a trivialization of the tangential bundle of $S^{n-1}$. This is true if and only if $n=2,4,8$ (the original reference is https://projecteuclid.org/euclid.bams/1183522319).
Now what about the group structure?

a) If $n=2$ then $S^{1}\times SO(1)\cong S^{1}$

b) If $n=3$, then $SO(4)\cong S^{3}\times SO(3)$, ~~in particular SO(4) is diffeomophic to a product of two Lie groups ~~ but it isn't a Lie groups isomorphism (see the comments below). (The Lie group structure on $S^{3}$ is the one give by the quaternions with norm $1$)

c) If $n=8$, then $SO(8)\cong S^{7}\times SO(7)$, but it is not longer a product of Lie group . $S^{7}$ may be viewed as the set of octonions with norm 1, it is in particular a quasi Lie group (or non associative Lie group).