For everything you're requesting, it seems reasonable only to consider this question using maps that are isomorphisms, since: (i) the zero map between semisimple Lie algebras has no reasonable "associated" map between root systems, (ii) inclusions such as ${\rm{Sp}}_{2n} \rightarrow {\rm{SL}}_{2n}$ have no reasonable associated map either way between the root systems (the latter is simply laced of rank $2n-1$ and the former has two root lengths and rank $n$).
After one restricts the morphisms under considerations to be isomorphisms (which is where all of the substance lies anyway), the answer is "yes". It comes down to a trick, but there is some real content underlying the trick. I have no idea what "strongly monoidal" means, but if one thinks about how mathematicians use semisimple Lie algebras, semisimple Lie groups, and semisimple algebraic groups (after they're done being classified) then such a classification that is functorial with respect to isomorphisms is not only elegant but also useful to do important things.
More to the point, the need for such a version of the classification theorem (really in a form with split connected semisimple groups over general fields in place of semisimple Lie algebras over $\mathbf{C}$) already arose back in the 1960's and especially 1970's, as part of the structure theory of connected semisimple groups over general fields (to classify Galois-twisted forms, especially over fields of arithmetic interest and over $\mathbf{R}$) and in the definition of the Langlands dual group beyond the split case. It is sometimes attributed to Kottwitz, but the main idea can already be found in SGA3 in the discussion of the "scheme of Dynkin diagrams" (see Exp. XXIV, 3.1--3.4) and it has probably been rediscovered multiple times (e.g., by Tits for his notion of $\ast$-action of Galois groups on diagrams when formulating classification theorems over general fields without split hypotheses).
So finally the question comes down to this: is there a way to associate a root system $\Phi(\mathfrak{g})$ to a complex semisimple Lie algebra $\mathfrak{g}$ in a manner that is functorial with respect to isomorphisms? Sure! Let $G = G(\mathfrak{g}) = {\rm{Aut}}_{\mathfrak{g}/\mathbf{C}}^0$ be the identity component of the automorphism scheme of $\mathfrak{g}$. It is a basic fact from the structure theory of connected semisimple groups and semisimple Lie algebras over $\mathbf{C}$ that the groups $G(\mathfrak{g})$ are exactly the connected semisimple algebraic groups with trivial center, and the natural map
$${\rm{Lie}}({\rm{Ad}}_G): {\rm{Lie}}(G) \rightarrow {\rm{Lie}}({\rm{GL}}(\mathfrak{g})) = {\rm{End}}(\mathfrak{g})$$
is an isomorphism onto $\mathfrak{g}$ (embedded via ${\rm{ad}}_{\mathfrak{g}}$). It is also a general fact that the natural action of $G(\mathbf{C})$ on $\mathfrak{g}$ is transitive on the set of pairs $(\mathfrak{h}, \mathfrak{b})$ consisting of Cartan subalgebras $\mathfrak{h} \subset \mathfrak{g}$ and Borel subalgebras $\mathfrak{b} \subset \mathfrak{g}$ containing $\mathfrak{h}$, and that the stabilizer in $G(\mathbf{C})$ of such a pair is the subgroup $T(\mathbf{C})$ where $T \subset G$ is the unique maximal torus whose Lie algebra is $\mathfrak{h}$ (inside $\mathfrak{g}$). For such a pair, let $(\Phi, \Delta)$ be the associated "based root datum" consisting of the root system for $(\mathfrak{g}, \mathfrak{h})$ and the root basis attached to $\mathfrak{b}$ (i.e., the simple roots in the positive system of roots consisting of those whose root lines are contained in $\mathfrak{b}$).
Note that the adjoint action on $T(\mathbf{C})$ on $\mathfrak{h}$ is trivial. Hence, if $(\mathfrak{h}', \mathfrak{b}')$ is another such pair in $\mathfrak{g}$, with $(\Phi', \Delta')$ the associated based root datum, then there is a canonical isomorphism $(\Phi, \Delta) \simeq (\Phi', \Delta')$: the adjoint action on $\mathfrak{g}$ arising from any element $g \in G(\mathbf{C})$ that carries $\mathfrak{h}$ onto $\mathfrak{h}'$ and carries $\mathfrak{b}$ onto $\mathfrak{b}'$ (such $g$ is unique modulo $T(\mathbf{C})$, so this isomorphism between based root data is independent of the choice of such $g$).
We define the canonical based root datum $(\Phi(\mathfrak{g}), \Delta(\mathfrak{g}))$ to be the inverse limit of all such $(\Phi, \Delta)$'s along the canonical isomorphisms just specified. More concretely, elements of $\Phi(\mathfrak{g})$ are exactly the compatible systems of roots with respect to those isomorphisms, and likewise for $\Delta(\mathfrak{g})$. It is clear from the construction that this pair is functorial with respect to isomorphisms among such $\mathfrak{g}$'s. Now compose this functorial construction (functorial with respect to isomorphisms) with the "forgetful" functor that drops the data of the root basis!
It is possible to have a non-trivial, closed, connected Lie subgroup $K$ of a compact, connected Lie group $G$ such that no element of $\mathfrak k$ is regular in $\mathfrak g$. For example, consider $K = \operatorname{SU}_2$ embedded in $G = \operatorname{SU}_4$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a &&& b \\ & a & b \\ & c & d \\ c &&& d \end{pmatrix}$.
On second thought, even this is overkill; you could just take $K$ to be the (non-trivial) central torus in $G = \operatorname U_2$. But maybe you wanted a non-central as well as non-trivial example.
Best Answer
Regular elements need not be semisimple! For example, in the Lie algebra $\frak{sl}_2$, every non-zero element is regular, with the centralizer spanned by the element itself. Among the elements of the standard basis, $e$ and $f$ are nilpotent, whereas $h$ is semisimple. Only $h$ spans a Cartan subalgebra; the subalgebras spanned by $e$ and $f$ are normalized by $h$, so they are not self-normalizing.
Your statement that a regular element belongs to a Cartan subalgebra is false, due to a missing crucial hypothesis: the element needs to be semisimple. Indeed, in a semisimple Lie algebra, a Cartan subalgebra is a maximal abelian subalgebra consisting of semisimple elements.