Let $(A, \le)$ be a lattice. Consider the following properties for a commutative operation $\cdot$ on $A$: $$c \cdot (a \wedge b) = (c \cdot a) \wedge (c \cdot b)$$ $$c \cdot (a \vee b) = (c \cdot a) \vee (c \cdot b)$$ $$a \cdot b = (a \wedge b) \cdot (a \vee b)$$ for all $a,b,c \in A$. (Actually, the third property comes from the first two, as pointed out by Xodarap in his answer below.)
For example, both $\wedge$ and $\vee$ satisfy the properties if and only if the lattice is distributive. Furthermore, if the lattice has a greatest (resp. least) element, then it is an identity for $\wedge$ (resp. $\vee$).
My question is the following:
What are some natural conditions on the lattice that guarantee the existence of an operation satisfying the properties above other than $\wedge$ and $\vee$? What other assumptions can we make (possibly on $\cdot$) to guarantee uniqueness?
I am mostly interested in associative operations with an identity.
The question arose from the problem of writing the sum and product of natural numbers in terms of the lattice operations respectively of $(\mathbb{N}, \le)$ and $(\mathbb{N}, \mid)$. In those cases I already know the existence of such operations, which are hidden in the definitions of the orders. Indeed, both orders are of the form $a \preceq b$ if and only if there exists $c$ such that $b = a*c$.
Actually, the construction of an order $\preceq$ like that can be carried out in any commutative monoid where the identity is irreducible and the cancellation law holds, but $\preceq$ is not necessarily a lattice order, unless $*$ enjoys a unique factorization property. Furthermore, uniqueness does not seem to follow from these premises only.
The first two hold true in any lattice-ordered group, and the third one is a consequence of the first two in any l-monoid:
Note that $(a\vee b)(a\wedge b)$ is equal to $(a^2\vee ab)\wedge(ab\vee b^2)$ - clearly $ab\leq (a^2\vee ab),(ab\vee b^2)$ so $ab$ is a lower bound of this product. An analogous one shows that $ab\geq (a\vee b)(a\wedge b)$ and therefore we conclude that $ab=(a\vee b)(a\wedge b)$.
Your last question is about uniqueness - I'm not sure exactly what you mean, but given that you can order monoids in a lot of different ways, I think you will need to put restrictions on both the ordering and the operation.