Forgive me my naiveté (and my spelling), but I think that on a small enough chart around $e$, we can actually subtract elements of $G$, and we have

$\displaystyle \left(\alpha\beta\right)^{\prime}\left(0\right)=\lim_{t\to 0}\underbrace{\frac{\alpha\left(t\right)\beta\left(t\right)-e}{t}}_{=\alpha\left(t\right)\cdot\frac{\beta\left(t\right)-e}{t}+\frac{\alpha\left(t\right)-e}{t}}$

$\displaystyle =\lim_{t\to 0}\left(\alpha\left(t\right)\cdot\frac{\beta\left(t\right)-e}{t}+\frac{\alpha\left(t\right)-e}{t}\right)$

$\displaystyle =\underbrace{\left(\lim_{t\to 0}\alpha\left(t\right)\right)}_{=\alpha\left(0\right)=e} \cdot \underbrace{\left(\lim_{t\to 0}\frac{\beta\left(t\right)-e}{t}\right)}_{=\beta^{\prime}\left(0\right)} + \underbrace{\left(\lim_{t\to 0} \frac{\alpha\left(t\right)-e}{t}\right)}_{=\alpha^{\prime}\left(0\right)}$

$\displaystyle =\beta^{\prime}\left(0\right)+\alpha^{\prime}\left(0\right) = \alpha^{\prime}\left(0\right)+\beta^{\prime}\left(0\right)$.

Note that I have used the product rule for limits ($\displaystyle \lim_{t\to 0}\left(U\left(t\right)\cdot V\left(t\right)\right)=\left(\lim_{t\to 0}U\left(t\right)\right)\cdot\left(\lim_{t\to 0}V\left(t\right)\right)$). This is okay because the multiplication on $G$ is continuous.

This looks a bit awkward, but as far as I know there is no other way to define the addition of tangent vectors than to use a chart, at least if tangent vectors are defined as stalks of curves. Thus, we cannot hope for a chart-less proof. I'd like to be proven wrong!