Two dimensional Lie Algebra - what do we know without knowing the Bracket?

For every field $K$ there are just two different $2$-dimensional Lie algebras. The first one is $K^2$, which is abelian (which means the bracket is zero). The second one is solvable, non-nilpotent, hence non-abelian, and the bracket of the basis $(x,y)$ is an arbitrary nontrivial linear combination of $x$ and $y$, i.e., $[x,y]=\alpha x+\beta y$ for some $\alpha,\beta\in K$ with $(\alpha,\beta)\neq (0,0)$. This Lie algebra is usually denoted by $\mathfrak{r}_2(K)$. All choices $(\alpha,\beta)\neq (0,0)$ give isomorphic Lie algebras, i.e., they are all isomorphic to $\mathfrak{r}_2(K)$. You may choose $[x,y]=x$. Then the adjoint operator $ad(x)$ given by $ad(x)(y)=[x,y]$ is not nilpotent, because one eigenvalue is equal to $1$. Hence $\mathfrak{r}_2(K)$ is not nilpotent by Engel's theroem. Also, it has only inner derivations, i.e. $Der(\mathfrak{r}_2(K))=ad (\mathfrak{r}_2(K))$. Its center is trivial. This gives another proof that it is not nilpotent, because nontrivial nilpotent Lie algebras have a nontrivial center.

The algebra $L=\mathfrak{r}_2(K)$ is solvable, since we have $$ [[L,L],[L,L]]=0. $$ It cannot be simple, since $I=\langle x \rangle$ is a proper Lie ideal.