I happened to notice that $2 \cdot 3 = 1+5$ and $2+3 = 1 \cdot 5$. I researched a little further:
With size (cardinality) 1 there is trivially an infinite amount of solutions.
With size 2 there is only the trivial $2+2 = 2\cdot 2$ in addition to the first one.
With size 3 there are at least $1+2+5=1 \cdot 1 \cdot 8$, $1+3+3=1\cdot 1\cdot 7$ and $2+2+2=1\cdot 1\cdot 6$.
The multisets could also be of different size: a trivial example would be $2 \cdot 2=4$ but bigger multisets would be more interesting because there would actually be addition and multiplication...
It is obvious that we can use brute force - the possible numbers are quite limited. But is there an elegant way of finding the amount of solutions for given sizes of $A$ and $B$? Of course the order of members doesn't matter, only the contents of the multisets.
Addendum on 3 Aug:
For exercise, I add some definitions and notions I encountered. But it's still not an answer.
Let $A$ and $B$ be multisets of $\mathbb{Z}_+$. We can define a relation:
$sprod(A, B): (\prod{A} = \sum{B}) \land (\prod{B} = \sum{A})$
(quite informal, but should suffice here)
This relation is not reflexive nor irreflexive (some of the multisets involved are sprod-related to themselves, most are not). The relation is symmetric: $sprod(A, B)$ implies $sprod(B, A)$.
For a multiset $A$ we can define an ordered pair $(s(A) = \sum{A}, p(A) = \prod{A})$.
$sprod(A, B)$ means that $s(A) = p(B)$ and $s(B) = p(A)$. So the brute force search must find multisets of given size where $p(A)\le s(A)$ because they are the rare cases.
If $p(A)=s(A)$ we have found $sprod(A, A)$. That could be called $selfsprod$ and indicated by $sprod(A, s(A))$. For example, $sprod(\{1,2,3\}, 6)$.
If $p(A)<s(A)$, there is a good chance that we find a mirror case $B$ where the ordered pair is swapped. Then $sprod(A, B)$.
If $n$ is prime, only $sprod(\{n\}, n)$. That is the trivial case that gives infinite solutions, so not any of our interest. There are no $sprod(A, n)$ so that $|A|>1$.
If $n$ is composite, there is always a selfsprod found: take at least one divisor out, write $n$ in that form (two or more divisors) and fill enough 1's so that the sum goes to $n$. For example, $sprod(\{1,1,1,3,3\}, 9)$. This also yields a family of sprods: if we write the previous in the form $sprod(\{1^3,3,3\}, 9)$, there are also $sprod(\{1^2,3,3\}, \{1,8\})$, $sprod(\{1,3,3\}, \{1,1,7\})$ and $sprod(\{3,3\}, \{1,1,1,6\})$. More generally, $sprod(\{1^n,A\}, k)$ gives $sprod(\{1^{n-1},A\}, \{1,k-1\})$ and all the way to $sprod(\{A\}, \{1^n,k-n\})$.
If $sprod(A, A)$, and so $sprod(A, n)$, in any sprod $A$ can be replaced with $n$ and vice versa. So, if there are 1,2 and 3, they can be replaced by 6, and 2 and 2 can be replaced by 4.
Don't worry, I won't accept this answer. It isn't elegant as I meant it. But it's quite conclusive and probably I won't get any further.
Algorithm. Handle composite numbers $n$ in order, until the required length is exhausted. As shown in the addendum, write $n$ as its divisors and fill with 1's so that also the sum reaches $n$. There are found a selfsprod $sprod(A,A)$ and also $sprod(A,n)$. The latter yields more sprods: as long as there are 1's on the left side, they can be moved to the right side and each time $n$ is decreased. Whenever there is a composite in the multiset, it can be replaced by the multiset obtained at that composite. This also pertains to $n$ itself: all its different representations are mutually sprod.
Theorem. The algorithm is complete: it gives every possible sprod pair.
Definitions. If $A$ is the kind of multiset handled here, its $payload A_{>1}$ is its subset that has exactly those elements differing from 1. So, if $A = \{1,1,2,2,2\}$, $A_{>1} = \{2,2,2\}$.
If $sprod(A,n)$, $|A|$ is sprod expansion factor, $sef$ for that expansion. The minimal of these comes from dividing $n$ by the smallest factor and constructing the multiset from that. It can be called $sef(n)$. So, $sef(6) =3$ because $sprod(\{1,2,3\},6)$.
Proof. Take multisets $A$ and $B$ that are supposed to be related: $sprod(A,B)$. Now we are only interested in their payloads, because the 1's follow unambiguously.
If $\sum{A_{>1}} > \prod{B}$, there is no way to make the relation. Likewise if $\sum{B_{>1}} > \prod{A}$.
We can assume that $\prod{A} > \prod{B}$.
If they were equal, we would have found a pair of representations directly from $\prod{A}$. For example, if $A_{>1} = \{2,6\}$ and $B_{>1} = \{3,4\}$, they can both be found from 12 in the list below.
And if $\prod{A} < \prod{B}$, we simply swap them.
Now start with $sprod(\{1^{\prod{A}-\sum{A_{>1}}}, A_{>1}\}, \prod{A})$.
Take $\prod{A}-\prod{B}$ steps and end up with
$sprod(\{1^{\prod{B}-\sum{A_{>1}}}, A_{>1}\}, \{1^{\prod{A}-\prod{B}}, \prod{B}\})$.
Since $\sum{A_{>1}} \le \prod{B}$, this is well-defined: there are always enough 1's to move.
This is given by the algorithm starting at $\prod{A}$. And $\prod{B}$ is replaced with $B$. $\blacksquare$
For example, $A_{>1} = \{7, 53\}$, $B_{>1} = \{3, 37\}$.
We start with $sprod(\{1^{311}, 7, 53\}, 371)$
and end up with $sprod(\{1^{51}, 7, 53\}, \{1^{260}, 111)$.
As $sprod(\{1^{71}, 3, 37\}, 111)$, the complete relation is
$sprod(\{1^{51}, 7, 53\}, \{1^{331}, 3, 37\})$.
And it was found from composite 371, making 260 transformations and finally replacing 111 with its sprod expansion.
Demonstration of the algorithm to total cardinality of 10
4:
$\{2,2\}, 4$
6:
$\{1,2,3\}, 6$
$\{2,3\}, \{1,5\}$
8a:
$\{1,1,2,4\}, 8$
$\{1,2,4\}, \{1,7\}$
$\{2,4\}, \{1,1,6\}$
8b:
There is $\{1,1,2,2,2\}, 8$ but it is obtained from 8a by replacing 4 with its expansion. So no need to handle separately.
9:
$\{1,1,1,3,3\}, 9$
$\{1,1,3,3\}, \{1,8\}$
$\{1,3,3\}, \{1,1,7\}$
$\{3,3\}, \{1,1,1,6\}$
10:
$\{1,1,1,2,5\}, 10$
$\{1,1,2,5\}, \{1,9\}$
$\{1,2,5\}, \{1,1,8\}$
$\{2,5\}, \{1,1,1,7\}$
12a:
$\{1,1,1,1,2,6\}, 12$ - note that from this point the selfsprods already exceed total 10 $\{1,1,1,2,6\}, \{1,11\}$
$\{1,1,2,6\}, \{1,1,10\}$
$\{1,2,6\}, \{1,1,1,9\}$
$\{2,6\}, \{1,1,1,1,8\}$
12b:
$\{1,1,1,1,1,3,4\}, 12$
$\{1,1,1,1,3,4\}, \{1,11\}$
$\{1,1,1,3,4\}, \{1,1,10\}$
$\{1,1,3,4\}, \{1,1,1,9\}$
$\{1,3,4\}, \{1,1,1,1,8\}$
$\{3,4\}, \{1,1,1,1,1,7\}$
12c:
$\{1,1,1,1,1,2,2,3\}, 12$ is obtained from 12b by replacing 4 with $\{2,2\}$.
14:
$\{1,1,1,1,1,2,7\}, 14$
$\{1,1,1,1,2,7\}, \{1,13\}$
$\{1,1,1,2,7\}, \{1,1,12\}$
$\{1,1,2,7\}, \{1,1,1,11\}$
$\{1,2,7\}, \{1,1,1,1,10\}$
$\{2,7\}, \{1,1,1,1,1,9\}$
15:
$\{1^7,3,5\}, 15$
$\{1^6,3,5\}, \{1,14\}$
$\{1^5,3,5\}, \{1^2,13\}$
$\{1^4,3,5\}, \{1^3,12\}$
$\{1^3,3,5\}, \{1^4,11\}$
$\{1^2,3,5\}, \{1^5,10\}$
$\{1,3,5\}, \{1^6,9\}$
$\{3,5\}, \{1^7,8\}$
16a:
$\{1^6,2,8\}, 16$
$\{1^5,2,8\}, \{1,15\}$
$\{1^4,2,8\}, \{1^2,14\}$
$\{1^3,2,8\}, \{1^3,13\}$
$\{1^2,2,8\}, \{1^4,12\}$
$\{1,2,8\}, \{1^5,11\}$
$\{2,8\}, \{1^6,10\}$
16b:
$\{1^8,4,4\}, 16$ - this already goes over the limit of 10, obviously so do also all its expansions
18a:
$\{1^7,2,9\}, 18$
$\{1^6,2,9\}, \{1,17\}$
$\{1^5,2,9\}, \{1^2,16\}$
$\{1^4,2,9\}, \{1^3,15\}$
$\{1^3,2,9\}, \{1^4,14\}$
$\{1^2,2,9\}, \{1^5,13\}$
$\{1,2,9\}, \{1^6,12\}$
$\{2,9\}, \{1^7,11\}$
18b:
$\{1^9,3,6\}, 18$ goes over 10
20:
$\{1^8,2,10\}, 20$ goes over 10, so will also the later ones, so we stop here
Actual sprods picked from above
Here are only those sprod-relations where the cardinalities are equal. In the similar way also other kinds can be extracted.
2:
$\{2,2\}, \{2,2\}$ from 4, expands 4 to selfsprod
$\{2,3\}, \{1,5\}$ from 6, one move
3:
$\{1,2,3\}, \{1,2,3\}$ from 6, expands 6 to selfsprod
$\{2,2,2\}, \{1,1,6\}$ from 8a, two moves, expands 4 to 2,2
$\{1,3,3\}, \{1,1,7\}$ from 9, two moves
$\{1,2,5\}, \{1,1,8\}$ from 10, two moves
4:
$\{1,1,2,4\}, \{1,1,2,4\}$ from 8a, expands 8 to selfsprod
$\{1,1,3,4\}, \{1,1,1,9\}$ from 12b, three moves
$\{1,1,2,7\}, \{1,1,1,11\}$ from 14, three moves
5:
$\{1,1,2,2,2\}, \{1,1,2,2,2\}$ from 8b, expands 8 to selfsprod
$\{1,1,1,3,3\}, \{1,1,1,3,3\}$ from 9, expands 9 to selfsprod
$\{1,1,1,2,5\}, \{1,1,1,2,5\}$ from 10, expands 10 to selfsprod
$\{1^3,3,5\}, \{1^4,11\}$ from 15, four moves
$\{1^3,2,9\}, \{1^4,14\}$ from 18a, four moves