Let $(L, ∨, ∧)$ be a lattice with infinite cardinality $\mathrm{Card}\,(L)=\alpha$, and Put
$$ S:=\{ f| f:L\times L\rightarrow L \mbox{ is a bijection and } x ∧ y\leq f(x,y)\leq x ∨ y, \mbox{ for all } x\neq y\} $$ Then:
(1) Is it true that $S\neq \emptyset$?
(2) What is $\mathrm{Card}\,(S)$ (if we accept the generalised continuum hypothesis)?
(3) In particular, if $L=\mathbb{R}$ (the lattice of real numbers with the usual order), then can we say $\mathrm{Card}\,(S)=2^c$ ?
Let $L=\mathbb{Z}$ with its usual order. If $f\in S$, then for any $n\in\mathbb{Z}$, $f(n,n+1)$ and $f(n+1,n)$ must each be either $n$ or $n+1$. It follows that $f$ is already surjective when restricted to the set of pairs of the form $(n,n+1)$ or $(n+1,n)$, and thus cannot be injective on all of $\mathbb{Z}\times \mathbb{Z}$. So $S$ is empty in this case.
On the other hand, suppose $L$ has the property that for any $x\neq y$, $|\{z:x\wedge y\leq z\leq x\vee y\}|=\alpha$ Then $S$ is nonempty and has cardinality $2^\alpha$. Indeed, you can construct elements of $S$ by a transfinite induction of length $\alpha$: enumerate $L\times L$ with order type $\alpha$, and define $f$ on each pair by induction such that $f$ is injective and $x \wedge y\leq f(x,y)\leq x \vee y$ whenever $x\neq y$. This is possible since the set of possible values satisfying this inequality has cardinality $\alpha$, and at any stage of the induction we have defined fewer than $\alpha$ different values of $f$. On pairs of the form $(x,x)$, you can define $f$ in such a way to make sure that $f$ is surjective, since there are $\alpha$ such pairs. Moreover, at each step of this induction (except the ones where we are ensuring $f$ is surjective) there are at least $2$ (in fact, $\alpha$ many) different choices of how to define $f(x,y)$, so this gives $2^\alpha$ different choices for $f$.
In particular, this applies to $L=\mathbb{R}$, to answer your third question.