The kernel of $\operatorname{ad}_x = [x, \cdot]$ consists of 0 and ${\rm span}\{x\}$, when is this true? The Lie algebra $\mathfrak{s}\mathfrak{o}(3)$, when identified as vectors in $\mathbb{R}^3$ with the cross product as the bracket, has the property that the only $y \in \mathfrak{s}\mathfrak{o}(3)$ such that $x \times y = 0$ are $y=0$ or $y = \lambda x$. This is just from geometric properties of the cross product as the unique (up to sign and magnitude) perpendicular line between the two vectors, and only returns 0 in the case they are both parallel or one vector is 0.
Is this something that holds in general Lie algebras? I have tried searching about the kernel of $\operatorname{ad}_x$ for a fixed $x$, and I've seen a couple of statements about the centre of the Lie group or Lie algebra but nothing as concrete as the case for $\mathfrak{s}\mathfrak{o}(3)$. What kind of Lie algebras have this property other than $\mathfrak{s}\mathfrak{o}(3)$?
 A: The concept you are looking for and want to read up on is that of regular elements of Lie algebras. See e.g. https://en.wikipedia.org/wiki/Regular_element_of_a_Lie_algebra.
Roughly speaking, those are the elements $x \in L$ whose centralizer $C_L(x)$, which is identical to $\ker(ad_x)$ is "as small as possible". Note that unfortunately, the nomenclature is a bit vague and might differ a little in some sources (e.g. often only semisimple elements in semisimple Lie algebras are considered etc.)
For semisimple Lie algebras $\mathfrak g$, that "minimal possible dimension" is shown to equal the dimension of any Cartan subalgebra of $\mathfrak g$ and is called the rank of $\mathfrak g$.
In other words, your only chance among semisimple $\mathfrak g$ to find more examples of what you are looking for, namely that minimal dimension to be $1$, is in rank $1$ Lie algebras.
Over $\mathbb R$ or $\mathbb C$, the only rank $1$ semisimple Lie algebras besides the (real) $\mathfrak{so}(3) = \mathfrak{su}_2$ are the real $\mathfrak{sl}_2(\mathbb R)$ and the complex $\mathfrak{sl}_2(\mathbb C)$. I leave it to you to check those.
And I leave the case of solvable Lie algebras to someone else.
