Tangent bundle of smooth closed simply-connected $4$-manifold $w_1 = w_2 = 0$ can be trivialized in complement of point?

Yes. For manifolds of dimension $\le 7$ the obstructions to trivializing any real vector bundle are $w_1, w_2$, and a class $\frac{p_1}{2} \in H^4(BSpin(n), \mathbb{Z})$ ($n \ge 3$), the fractional first Pontryagin class, defined in the presence of a spin structure. On an open $4$-manifold $H^4(-, \mathbb{Z})$, and hence $\frac{p_1}{2}$, vanishes.

$M$ is smooth, so there is a cellular decomposition with a single 4-cell. Thus we want $TM$ trivializable over the 3-skeleton, so we look to analyze sections of the frame bundle which has Lie group fiber $SO(4)$ (here $M$ is orientable because $w_1=0$). Obstruction theory says we can extend over the 2-skeleton, because the class is $w_2\in H^2(M; \pi_1SO(4)=\mathbb{Z}_2)$ and you assumed that to be zero. The next obstruction (extending over the desired 3-skeleton) is automatically vacuous, because $\pi_2SO(4)=0$.