Let $G$ be a Lie group. There is a diffeomorphism $$G \times T_e G \to TG$$ mapping $(g, [\gamma]) \mapsto [g \cdot \gamma]$. The inverse map then gives rise to the following isomorphism of bundles: $$TG \to G \times \mathbb{R}^n$$ mapping $[\varphi] \mapsto (\varphi(0), (x(\varphi(0)^{-1}\cdot\varphi))'(0))$ where $\varphi(0)^{-1}$ denotes the inverse element of $\varphi(0)$ w.r.t to the binary operation on the group and $x$ is a chart about the idenity. This is bijective, smooth, and linear on the fibers, as it is simply a composite of the (inverse map of the) diffeomorphism given above, followed by an inclusion in the first coordinate and the known isomorphism of vector spaces $$T_e G \cong \mathbb{R}^n$$ in the second.
Does the argument above seem sound?
Yes, this is correct.
Another way of proving this fact is to pick an ordered basis $\{v_1, \dots, v_n\}$ for $T_eG$ and define vector fields $V_1, \dots, V_n$ by $V_i(g) = dL_g(v_i)$ where $L_g : G \to G$ is given by $v \mapsto gv$. As $dL_g$ is an isomorphism ($(dL_g)^{-1} = dL_{g^{-1}}$), $\{V_i(g) \mid i =1, \dots, n\}$ is a basis of $T_gG$ for every $g \in G$. Therefore $TG$ is parallelisable.
Denote the isomorphism $T_eG \to \mathbb{R}^n$ and the diffeomorphism $G\times T_eG \to TG$ in your answer by $\varphi$ and $\psi$ respectively. The relationship between your proof and the vector fields constructed above is given by the identity $V_i(g) = \psi(g, \varphi^{-1}(e_i))$; this identity holds under the assumption that the ordered basis $\{v_1, \dots, v_n\}$ chosen in the previous paragraph is the same as the one chosen in order to define the isomorphism $\varphi$ (i.e. $\varphi(v_i) = e_i)$.