I am currently dealing with Lie groups. To show that the tangent space at the neutral element of a Lie group carries the structure of a Lie algebra I tried to prove the following Lemma:
Let G be a Lie group of dimension n.
(i) For every tanget vector $\xi \in T_{e}G$ there exists exactly one left-translation-invariant vector field $X^{\xi} \in \Gamma^{\infty}(TG)$ with $X^{\xi}(e)=\xi$. It is given by
\begin{equation}
X^{\xi}(g) = d_{e}L_{g}(\xi)
\end{equation}
where $L_{g}:G \rightarrow G$ is defined as $L_{g}(x)=gx$. In particular, the mapping $\xi \mapsto X^{\xi}$ is linear.
(ii) The Lie bracket of two left-translation-invariant vector fields is again left-translation-invariant.
For left-translation-invariant vector fields we have the following definition:
Let $G$ be a Lie group. A vector field $X \in \Gamma^{\infty}(TG)$ is said to be left-translation-invariant, if \begin{equation} d_{p}L_{g}X(p) = X(L_{g}(p)) \quad \text{for all }p,g \in G \end{equation} In other words: $X$ is $L_{g}$-related to itself.
I came up with the following proof:
(i) To show the existence of such a vector field, it is sufficient to show that $X^{\xi}(g)=d_{e}L_{g}(\xi)$ is a left-translation-invariant vector field with the property $X^{\xi}(e)=\xi$. It is
\begin{equation} X^{\xi}(e) = d_{e}L_{e}(\xi)=\xi \end{equation}
Furthermore it holds
\begin{align} d_{p}L_{g}X^{\xi}(p) &= d_{p}L_{g}(d_{e}L_{p}(\xi)) \\
&= d_{e}(L_{g}\circ L_{p})(\xi) \\
&= d_{e}L_{gp}(\xi) \\
&= X^{\xi}(gp) \\
&= X^{\xi}(L_{g}(p)) \end{align}
So $X^{\xi}$ is such a vector field. To show the uniqueness, let $Y$ be another left- translation-invariant vector field with $Y(e)=\xi$. Consider
\begin{align}
X^{\xi}(p)-Y(p) &= d_{e}L_{p}(X^{\xi}(e))-d_{e}L_{p}(Y(e)) \\
&= d_{e}L_{p}(X^{\xi}(e)-Y(e)) \\
&= d_{e}L_{p}(\xi-\xi) \\
&= d_{e}L_{p}(0) \\
&= 0
\end{align}
The linearity of the map $\xi \mapsto X^{\xi}$ follows directly from the linearity of the differential:
\begin{equation} X^{\lambda\xi+\eta}(g) = d_{e}L_{g}(\lambda\xi+\eta) = \lambda d_{e}L_{g}(\xi) + d_{e}L_{g}(\eta) = \lambda X^{\xi}(g)+X^{\eta}(g) \end{equation}
This proves (i).
(ii) follows directly from following statement on Lie brackets
Let $F:M \rightarrow N$ be a smooth map between manifolds and let $X_{1}, X_{2} \in \Gamma^{\infty}(TM)$ and $Y_{1}, Y_{2} \in \Gamma^{\infty}(TN)$ be vector fields such that $X_{i}$ is F-related to $Y_{i}$ for $i=1,2$. Then $[X_{1},X_{2}]$ is F-related to $[Y_{1},Y_{2}]$.
Although I think the proof is correct, I would be grateful if someone would look at it again.
I can't use comments yet, therefore I write this answer.
I have corrected some typos in Your post but other than that it seems to be ok. The calculations look good and if I am not missing anything Your reasoning seems to be correct too.