Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology?

The answer to the question is positive, due to Wu's formula. See e.g. Milnor-Stasheff, Characteristic classes, lemma 11.13 and theorem 11.14. In fact, all one needs to compute the Stiefel-Whitney classes of a smooth compact manifold (orientable or not) is the cohomology mod 2 (as an algebra) and the action of the Steenrod algebra on it. Both structures are preserved under cohomology isomorphisms induced by continuous maps.

The most conceptual way of understanding the relation between the mod 2 Wu and Stiefel-Whitney classes of a manifold and the action of the Steenrod algebra on the mod 2 cohomology is to use the homotopy theory of Poincare duality spaces and the Spivak normal fibration, and also the chain homotopy theory of chain complexes with symmetric Poincare complexes and the normal chain bundle expounded in my 1980 paper The algebraic theory of surgery (Part I, Part II). A map $f:L\to N$ of $n$-dimensional manifolds which induces isomorphisms in $Z_2$-coefficient cohomology also induces a chain equivalence of $n$-dimensional symmetric Poincare complexes over $Z_2$. Such a chain equivalence automatically preserves the Spivak normal chain bundles. The mod 2 Wu and Stiefel-Whitney classes of the manifolds are preserved by $f$ because they only depend only on the underlying chain homotopy structure. It is also worth reminding ourselves that Atiyah's 1960 paper Thom complexes established the $S$-duality between the Thom space of the normal bundle of a manifold $X$ and $X_+=X \cup \{*\}$, and so proved a conjecture of Milnor and Spanier: the stable fibre homotopy type of the tangent sphere bundle of a differentiable manifold $X$ depends only on the homotopy type of $X$.