If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic

The result you are hoping for is in fact false.

In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent bundle of $U\times\mathbb{R}^k \subset \mathbb{R}^{m+k}$ is trivial, while in Corollary 9.3, Milnor shows that it admits a smooth structure for which the tangent bundle has a non-zero Pontryagin class. As Pontryagin classes are stable, the stable tangent bundle of the latter manifold is not trivial, and hence not isomorphic to the stable tangent bundle of $U\times\mathbb{R}^k$ with its standard smooth structure.

Milnor, John W., Microbundles, Topology 3, Suppl. 1, 53-80 (1964). ZBL0124.38404.

Let me add something to Michael Albanese's great answer to see this question in a broader context.

Novikov proved that the rational Pontryagin classes are homeomorphism invariants (in fact, this was one of the achievements for which he received the Fields medal in 1970). The integral Pontryagin classes, however, are not invariant under homeomorphism, see Chapter 4.4 of "The Novikov Conjecture" by Kreck and Lück.

Some polynomials in the Pontryagin classes are even homotopy invariant: for instance, $p_1$ of a closed oriented $4$-manifold $M$ agrees (by Hirzebruch's signature theorem) with $3\sigma(M)$ times the fundamental class in cohomology (where $\sigma(M)$ denotes the signature of $M$), which is invariant under homotopy equivalence.

The famous Novikov conjecture asks whether certain so-called "higher signatures" are also invariant under homotopy equivalence. It is one of the most important open questions in topology.