Re 1): By "Replacing $y$ by $a^{-1}y$", the author means you let $z = a^{-1}y$, and then take the basis $(x,z)$ instead of $(x,y)$. Then you have $$[xz] = [x(a^{-1}y)] = a^{-1}[xy] = a^{-1}(ax) = x.$$
Re 2): $[xy] = x$ means you define the product for the two basis elements, in general you have
$$[(ax+by)(cx+dy)] = [(ax)(cx+dy)] + [(by)(cx+dy)] = ac[xx]+ ad[xy] + bc[yx] + bd[yy]$$
and use $[xx] = 0 = [yy]$ and $[yx] = -[xy]$.
(1) Yes, this is correct. One can cut the amount of work by $2/3$ if one observes that the Jacobian identity is skew-symmetric in $ijk$. If we denote the full alternation over these indices by $\textrm{Alt}$, the Jacobi identity is $$0 = \textrm{Alt} \,[x_i, [x_j, x_k]] ,$$
which gives
$$0 = \text{Alt} (a^l_{ij} a^m_{kl}) ;$$
optionally, we can write out the full alternation (and use antisymmetry of the lower indices of structure constants) to recover your formula.
(2) This is the strongest result possible, in the sense that (again, given the antisymmetry in the lower induces of the structure constants) this is equivalent to the Jacobi identity for any (equivalently, every) choice $(x_i)$ of basis.
(3) There isn't exactly an analogue of those statements---in both of those cases, one fixes the upper index $l$ of $a^l_{ij}$ and interprets the entries $a^l_{ij}$ as components of a matrix $A$, but in the Jacobi identity condition one contracts with this index, so it cannot be fixed in any sense.
Here are two ways we might see this a little more concretely, however:
(1) Using antisymmetry, we can rearrange the Jacobi identity as
$$[x, [y, z]] = [[x, y], z] + [y, [x, z]] .$$ Now, if we define for $x \in L$ the operator $\textrm{ad}_x : L \mapsto L$, $y \mapsto [x, y]$, we can rewrite this as
$$\textrm{ad}_x[y, z] = [\textrm{ad}_x(y), z] + [y, \textrm{ad}_x(z)] .$$ In other words, the Jacobi identity is just a Leibniz (product) rule for the operator $\textrm{ad}_x$.
(2) In the special case $L = \mathfrak{so}(3, \Bbb R) \cong \mathfrak{su}(2)$, we may identify the Lie bracket with the usual cross product $\times$ on $\Bbb R^3$. Then, the Jacobi identity translates to the possibly familiar identity
$${\bf x} \times ({\bf y} \times {\bf z}) + {\bf y} \times ({\bf z} \times {\bf x}) + {\bf z} \times ({\bf x} \times {\bf y}) = 0.$$
Best Answer
A typical element of $L$ is $u=\sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=\sum_j d_j x_j$. Then $$[u,v]=\sum_{i,j}c_id_j[x_i,x_j] =\sum_k\left(\sum_{i,j}a_{i,j}^kc_id_j\right)x_k.$$ This is nothing more than the bi-linearity of the Lie bracket.