Your second property is simply inconsistent with the nature
of a pairing function, since we want $p(n,m)=p(n',m')\iff
n=n'$ and $m=m'$. That is, a pairing function must be
one-to-one on pairs, but multiplication is not, since
$2\cdot 6=3\cdot 4$.

The first property, however, is easy to arrange as follows
(and there will be continuum many different such
functions). Let me work only with positive integers.
First, list all the possible product values in order.
For each such product $r$, observe that there are only
finitely many pairs with $nm=r$; list these pairs in any
desired order. Now, concatenate these lists of pairs, and
let $p(n,m)$ be the place of the pair $\langle n,m\rangle$
in your master list. This is a pairing function, since it
is injective on pairs, and it is monotone with respect to
products, since larger products appear later on the master
list.

Finally, note that it is not possible to achieve the property if you allow $0$, since in this case there are infinitely many pairs with product $0=n\cdot 0$, and they cannot all have pairing value before the the other pairs.

**Edit.** In your comments, you drop the one-to-one requirement, which makes this very far from what would ordinarily be called a pairing function. Nevertheless, the problem now admits the following rather silly solution in the positive integers: let $p(n,m)=nm$, which has both your stated properties. The "inverse" function is: let $F(r)=r$ and $G(r)=1$. That is, given the product $r$, we return the pair $\langle r,1\rangle$, which of course also has product $r\cdot 1=r$. In otherwords $p(F(r),G(r))=r$, which would seem to be the inverse requirements.