One can ask
When is $T^1 S^n$, the unit tangent bundle of $S^n$, abstractly diffeomorphic to $S^{n-1}\times S^n$?
For even $n$, the answer is never. This is because $T^1 S^{2n}$ has torsion in its cohomology ring, but $S^{2n-1}\times S^{2n}$ doesn't, so $T^1 S^{2n}$ is not even homotopy equivalent to $S^{2n-1}\times S^{2n}$.
For odd $n$, the answer is more delicate.
Since $S^1$, $S^3$, and $S^7$ are parallelizable, $T^1 S^1$, the unit tangent bundle to $S^1$, is diffeomoprhic to $S^0\times S^1$. Likewise, $T^1 S^3$ is diffeomorphic to $S^2\times S^3$. The same argument shows $T^1 S^7$ is diffeomorphic to $ S^6\times S^7$. In all of these cases, a much stronger statement is true: $T^1 S^{k}\cong S^{k-1}\times S^k$ as bundles over $S^k$ (for $k=1,3,7$.)
So, the first case where I don't know the answer is $T^1 S^5$. The first thing to say is that $S^5$ is not parallelizable, so $T^1 S^5$ is not bundle isomorphic to $S^4\times S^5$. I can prove that $T^1 S^5$ and $S^4\times S^5$ have the same cohomology rings and that their respective tangent bundles have the same Stiefel-Whitney, Pontrjagin, and Euler classes. So none of the "usual" invariants distinguish them. (More generally, I think I can prove that $T^1 S^{2n-1}$ and $S^{2n-2}\times S^{2n-1}$ have isomorphic cohomology rings and the same characteristic classes).
On the other hand, a paper by De Sapio and Walschap, "Diffeomorphism of total space and equivalence of bundles" proves that $TS^n$ and $\mathbb{R}^n\times S^n$ are not abstractly diffeomorphic, unless they are also bundle isomorphic. So, we know that $TS^5$ is not diffeomoprhic to $\mathbb{R}^5\times S^5$. So, if $T^1 S^5$ and $S^4\times S^5$ are diffeomorphic, then no diffeomorphism can extend to a diffeomorphism of $TS^5$ and $\mathbb{R}^5\times S^5$.
I'm not sure how hard the answer for $S^5$ will compared to any other $S^{k}$ with $k$ odd $\neq 1,3,7$, so I'll ask the $S^5$ question separately:
Is $T^1 S^5$ abstractly diffeomorphic to $S^4\times S^5$?