Lie Group Structure on the 2-Sphere: does the following argument hold?

In principle, your approach is correct, but it leads to needless complex computations. I would like to suggest three other ways of proving that no Lie structure can be defined on $S^2$.

1. This is actually your own suggestion. Since on any $n$-dimensional Lie $G$ group one can define $n$ vector fields $X_1,X_2,\ldots,X_n$ which are linearly independent everywhere (that is, for each $g\in G$, the vectors $X_1(g),X_2(g),\ldots,X_n(g)$ are linearly independent), the hairy ball theorem allows us to deduce that no Lie group structure can be defined on $S^2$.
2. The space $S^2$ is compact. If it was a Lie group, its Lie algebra $\mathfrak g$ would then be reductive (this is a standard fact within Lie group theory and not hard to prove). But there are only two $2$-dimensional Lie algebras (up to isomorphism). The one spanned (as a vector space) by two elements $X$ and $Y$ such that $[X,Y]=Y$ is not reductive. The other one is the abelian $2$-dimensional Lie algebra, which is the Lie algebra of one and (up to isomorphism) only one simply-connected Lie group, which is $(\mathbb{R}^2,+)$. And $\mathbb{R}^2$ and $S^2$ are not homeomorphic.
3. A slightly different approach, compared with the previous one, consists in finding a simply-connected Lie group $G$ whose Lie algebra $\mathfrak g$ is the non-abelian one from the previous point. And that's easy. Take$$G=\left\{\begin{bmatrix}a&b\\0&\frac1a\end{bmatrix}\,\middle|\,a>0\wedge b\in\mathbb{R}\right\}.$$Again, it is not compact, and therefore $G$ and $S^2$ are not homeomorphic.

The nontrivial $2$-dimensional Lie algebra $\mathfrak{g}$ is nilpotent. It's a general feature of nilpotent Lie algebras that the exponential map $\mathfrak{g} \to G$ to the corresponding simply connected Lie group is a diffeomorphism: in particular, $G$ is necessarily noncompact, and hence cannot be diffeomorphic (or in fact even homotopy equivalent) to $S^2$.

This is a good idea for a proof (I don't think I've actually seen it before), but it doesn't generalize well: it's in fact the case that no sphere $S^n, n \ge 2$ admits the structure of a Lie group except $S^3$, but as $n$ increases this gets harder and harder to prove with Lie theory. By contrast, a generalization of the hairy ball theorem immediately proves that the even-dimensional spheres don't admit the structure of a Lie group.