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Assuming distributivity, the answer to the first part of Question 1 is simple:

Every finite distributive lattice $L$ admits a multiplication, namely the meet operation.

1. Meet is commutative and associative.
2. The top acts as the multiplicative identity: $1 \wedge a = a$ for all $a \in L$.
3. For all $a \in L$ and $B \subseteq L$, the condition $a \wedge \left(\bigvee_{b \in B} b\right) = \bigvee_{b \in B} (a \wedge b)$ follows from distributivity and finiteness.

But there may be also other multiplications. For example, the three-element chain $0 < 1 < 2$ admits two multiplications, whose multiplication tables are:

[0 0 0]  [0 0 0]
[0 0 1]  [0 1 1]
[0 1 2], [0 1 2]

(the latter is the meet).

The four-element chain $0 < 1 < 2 < 3$ admits six lattice multiplications:

[0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]
[0 0 0 1]  [0 0 0 1]  [0 0 0 1]  [0 0 1 1]  [0 1 1 1]  [0 1 1 1]
[0 0 0 2]  [0 0 1 2]  [0 0 2 2]  [0 1 2 2]  [0 1 1 2]  [0 1 2 2]
[0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3]

The four-element diamond $0 < 1,2 < 3$ admits only one lattice multiplication, namely the meet.

These were found with a relatively brute-force search. How to find all lattice multiplications efficiently, and what happens with nondistributive lattices, I do not know.