The "scary expression" will give you what you want, but you need to be careful with the names of the indices, you have some conflicts there, like $i$ appearing twice in not summed expressions.
Instead, let's streamline things a bit, for $A,B\in \mathfrak g$ we have $$ A=a^{ij}\partial_{ij}|_e,\ B=b^{ij}\partial_{ij}|_e $$ with the components being constants, and summing over repeated indices are understood. A generic group element is denoted as $x^{ij}$ and $\partial_{ij}$ are the holonomic frame vectors associated with the canonical coordinate system mapping group elements to their matrix elements.
We also recall that given any manifold $M$ with local chart $(U,\varphi)$ (with $\varphi(x)=(x^1(x),...,x^n(x))$), if two vector fields $X=X^i\partial_i$ and $Y=Y^i\partial_i$ are given, then their commutator is locally given by $$ [X,Y]=\left(X^j\partial_j Y^i-Y^j\partial_j X^i\right)\partial_i. $$
Left multiplication:
Let $\gamma:(-\epsilon,\epsilon)\rightarrow G$ be a smooth curve such that $\gamma(0)=x$ and $\dot\gamma(0)=X=X^{ij}\partial_{ij}|_x$. Let $g=(g^{ij})\in G$ be a group element. Then $$ (l_g)_\ast X=\frac{d}{dt}g\gamma(t)|_{t=0}=^!gX=g^{ik}X^{kj}\partial_{ij}|_{gx}, $$ where at the equality sign with the exclamation mark we use the fact that the group elements are just ordinary matrices embedded into $\mathbb R^{n\times n}$.
So left translation of vectors in $\text{GL}(n,\mathbb R)$ is just ordinary left multiplication of the vector (which is a matrix, rememeber!).
The derivation:
By the previous, the left invariant vector fields corresponding to $A,B\in\mathfrak g$ (also denoted the same way) are given by $$ A_x=x^{ik}a^{kj}\partial_{ij}|_x\ B_x=x^{ik}b^{kj}\partial_{ij}|_x. $$
If we now reinterpret the $x^{ij}$ from being specific variables to being coordinate functions, we can also write the vector fields without evaluation at a specific point as $$ A=x^{ik}a^{kj}\partial_{ij}\ B=x^{ik}b^{kj}\partial_{ij}. $$ the commutator is then $$ [A,B]=(A^{mn}\partial_{mn}B^{ij}-B^{mn}\partial_{mn}A^{ij})\partial_{ij}, $$ where $$ A^{ij}=x^{ik}a^{kj}, $$ and similarly for $B$. This is $$ [A,B]=\left( x^{mr}a^{rn}\partial_{mn}(x^{ik}b^{kj})-x^{mr}b^{rn}\partial_{mn}(x^{ik}a^{kj}) \right)\partial_{ij}=^!\left(x^{mr}a^{rn}\delta^i_m\delta^k_n b^{kj}-x^{mr}b^{rn}\delta^i_m\delta^k_n a^{kj}\right)\partial_{ij} \\ =\left( x^{ir}a^{rk}b^{kj}-x^{ir}b^{rk}a^{kj} \right)\partial_{ij}. $$ At the equality with the exclamation mark we have used that $\partial_{mn}x^{ij}=\delta^i_m\delta^j_n$ and that the coefficients $a^{ij},b^{ij}$ are constants.
For the identity element we have $x^{ij}(e)=\delta^{ij}$, so $$ [A,B]_e=\left(a^{ik}b^{kj}-b^{ik}a^{kj}\right)\partial_{ij}|_e=[A_e,B_e], $$ where the last expression is the ordinary matrix commutator of the matrices $A_e,B_e$.
Maybe you just lack familiarity with the notation. Let's write out the formula $[E_i, E_j] = \sum_k C_{ijk} E_k$ which seems to confuse you, for the instances $ (1,1), (1,2)$ and $(2,1)$ for $(i,j)$:
$$[E_1, E_1] = \underbrace{C_{111}}_{0} E_1 + \underbrace{C_{112}}_0 E_2 + \underbrace{C_{113}}_0 E_3 = 0$$
$$[E_1, E_2] = \underbrace{C_{121}}_{0} E_1 + \underbrace{C_{122}}_0 E_2 + \underbrace{C_{123}}_1 E_3 = E_3$$
$$[E_2, E_1] = \underbrace{C_{211}}_{0} E_1 + \underbrace{C_{212}}_0 E_2 + \underbrace{C_{213}}_{-1} E_3 = -E_3$$
etc. Surely you can fill in the other possibilities for $(i,j)$ and see how this is just a different notation for what you wrote.
Best Answer
Let $U$ be an open neighborhood of $e$ in $G$ such that the local coordinates system $x=(x^i)$ is well defined on $U\cdot U$. Given $g\in G$, denote $L_g$ the left translation by $g$, i.e. $$L_g:G\to G,\quad h\mapsto gh.$$ When $g,h\in U$, following your notations, $$x^k(L_g(h))=m(x(g),x(h))^k=x^k(g)+x^k(h)+\frac{1}{2}b_{i,j}^kx^i(g)x^j(h)+O_3\big(x(g),x(h)\big),$$ so for fixed $g$, the Jacobi matrix of the map $x(h)\mapsto x(L_g(h))$ at $h=e$ is given by $$\frac{\partial x^k(L_g(h))}{\partial x^j(h)}\Big|_{h=e}=\delta_j^k+\frac{1}{2}b_{i,j}^kx^i(g)+O_2(x(g)). \tag{1}$$ Let $(L_g)_*:TG\to TG$ be the tangent map of $L_g$, which maps $T_eG\to T_gG$. Since $e_j$ is a left invariant vector field with $e_j(e)=\partial_j(e)$, then by $(1)$, $$e_j(g)=(L_g)_*\partial_j(e)=\frac{\partial x^k(L_g(h))}{\partial x^j(h)}\Big|_{h=e}\cdot \partial_k(g)=\partial_j(g)+\frac{1}{2}b_{i,j}^kx^i(g)\cdot\partial_k(g)+O_2(x(g)).$$ It follows that at $g=e$,
$$ [e_i,e_j]=[\partial_i+\frac{1}{2}b_{k,i}^lx^k\partial_l\,,\partial_j+\frac{1}{2}b_{m,j}^nx^m\partial_n]=\frac{1}{2}\left(\partial_i(x^m)b_{m,j}^n\partial_n-\partial_j(x^k)b_{k,i}^l\partial_l\right)=c_{ij}^ke_k. $$