No. Take $A = \{ 2, 10, 9, 12 \}$ and $B = \{ 4, 5, 6, 18 \}$.
Here is how I constructed the above example. I found a number, namely $12$, which admits two factorizations $12 = 2 \cdot 6 = 3 \cdot 4$
such that the sum of the numbers used in each factorization differs by a small amount (in this case $1$). I found a second such number, namely $20$: $20 = 4 \cdot 5 = 2 \cdot 10.$
In this case the sum of the numbers used in each factorization differs by $3$. If I multiply each of the numbers in the first example by $3$, I get a second number $108 = 6 \cdot 18 = 9 \cdot 12$
with two factorizations such that the sum of the numbers used in each factorization differs by $3$. By mixing these I can arrange for the sums to be equal, and by construction the products are clearly equal.
Generalizing the above argument yields the $6$-parameter family of examples $A = \{ ta, tbc, sde, sf \}, B = \{ tab, tc, sd, sef \}$
where $t = (de + f) - (d + ef)$ and $s = (ab + c) - (a + bc)$. (I may have a sign wrong.) One obtains the above example by setting $a = 3, b = c = 2, d = 5, e = f = 2$.
Here is a counterexample with triplets: $A = \{ 2, 8, 9 \}, B = \{ 3, 4, 12 \}$. To get these I started with numbers of the form $A = \{ a, bc, de \}, B = \{ ab, cd, e \}$
which have the same product; demanding they have the same sum gives $a + bc + de = ab + cd + e.$
Setting $b = 2$ gives $a = e(d-1) - c(d-2)$
and yields a $3$-parameter family of solutions in the parameters $c, d, e$. The above example is obtained by setting $d = 3, c = 4, e = 3$.