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For a simple example with complete total orders, take $L=\{0\}\cup[1,2]$ and $K=\{-1,0\}\cup[1,2]$.

No; in fact, we can canonically recover $L$ from $\mathcal{Id}(L)$ as the sublattice of compact elements (that is, elements $x$ such that whenever $x=\bigvee S$, there is a finite subset $F\subseteq S …

There is a trivial upper bound of $2^\kappa$. In fact, this bound can already be achieved for complete totally ordered sets. For instance, given any subset $A\subseteq\kappa+1$, let $L_A$ be obtaine …

Here's a counterexample. Let $L=\{0,1,x_0,x_1,x_2,\dots,y\}$, where $x_0<x_1<x_2<\dots$ and $y$ is incomparable with every $x_n$. Then the only non-principal ideal in $L$ is $I=\{0,x_0,x_1,\dots\}$; …

Yes, this is apparently a fairly hard theorem of Pudlak and Tuma (or at least I assume it is hard, because it seems to have been an open problem for decades before they finally proved it in 1980).