Suppose $\{e_1, e_2, e_3\}$ are the standard ordered basis in the form of column vectors of the vector space $\mathbb{R}^3$ over the field $\mathbb{R}$. If $\mathcal{L}: \mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation defined by $$\begin{align} \mathcal{L}(e_1) &= e_1 \\ \mathcal{L}(e_2) &= e_1 + e_2 \\ \mathcal{L}(e_3) &= e_2 + e_3 \end{align}$$ show that $\mathcal{L}$ spans $\mathbb{R}^3$.
I attempted it this way. Since $\det [e_1 \quad e_1 + e_2 \quad e_2 + e_3] = 1 \neq 0$, the set $\{e_1, e_1+e_2, e_2+e_3\}$ must be linearly independent and hence, a basis of $\mathbb{R}^3$. Thus, for any $x \in \mathbb{R}^3$, $$\exists \ \alpha, \beta, \gamma \in \mathbb{R} : \alpha (e_1) + \beta (e_1 + e_2) + \gamma (e_2 + e_3) = x \implies \alpha e_1 + \beta e_2 + \gamma e_3 \stackrel{\mathcal{L}}{\mapsto} x.$$
I am looking for alternate proofs. In particular, I am interested in a proof that uses invertibility (more directly).
Note $\mathcal{L}$ admits an inverse $$\mathcal{L}^{-1}:\mathbb{R}^3\to\mathbb{R}^3,(x,y,z)\mapsto(x-y,y-z,z)$$ Which is also a linear map, thus $\mathcal{L}$ is an isomorphism, and it spans $\mathbb{R}^3$.