Let $k$ be an algebraically closed field of characteristic $0$. It is a well known fact that there is a finite-dimensional Lie algebra over $k$ which is not the Lie algebra of an affine algebraic $k$-group -- see https://mathoverflow.net/a/576. However, it is unclear to me whether or not every finite-dimensional Lie algebra $\mathfrak{g}$ over $k$ is the Lie algebra of a (not necessarily affine) algebraic group.
The question then is: is there a finite-dimensional Lie algebra over $k$ which is not the Lie algebra of an algebraic $k$ group? I have a strong impression this is the case, but I could only find examples of the statement regarding affine groups -- which is weaker than what I am asking for in here. If the answer to this question is "yes", does the statement remain true if we allow for group $k$-schemes in general? In other words, is every finite-dimensional Lie algebra the Lie algebra of a group scheme?
Motivation for this question: Given an algebraic $k$-group $G$, the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$ of its Lie algebra $\mathfrak{g}$ is canonically isomorphic to the algebra of $G$-invariant differential operators $k[G] \to k[G]$. This can be used to prove results regarding the structure of $\mathcal{U}(\mathfrak{g})$, such as the fact it is a domain and the Poincaré-Birkoff-Witt theorem (prove the fact that the theorem holds for this particular Lie algebra). Thus, if every finite-dimensional Lie algebra is the Lie algebra of a algebraic group, this isomorphism can be used to prove general statements about universal envoloping algebras of finite-dimensional algebras.
If $G$ is a connected algebraic group with trivial center, then it is affine (follows from Theorem 1 in Brion's paper https://arxiv.org/abs/1509.03059). Hence it is enough to find any non-algebraic Lie algebra with trivial center.
Example ($k$ algebraically closed of char zero): fix an irrational $s$ and consider $\mathfrak{g}(s)=k^2\rtimes k$, where $k$ acts by the derivation given by scalar multiples of the diagonal matrix $(1,s)$. This is non-algebraic (because it has weights that are irrationally collinear).