I have a solution, but not sure.... We can write $1, 2, 3, 4, 5, 6$ on the sides of one die and $1, 1, 1, 7, 7, 7$ on the sides of the other. Then each of the $12$ possible sums appears in exactly $3$ cases.

Hm, sure: $1,1,1,7,7,7$ and $1,2,3,4,5,6$ The probability of a number $x$ where $2 \le x \le 7$ is obtained from getting $1$ in the first die and a corresponding number in the second die. The probability of a number $x$ where $8 \le x \le 13$ is obtained from getting $7$ in the first die and a corresponding number in the second die. Since all numbers in the second die are equally likely and both $1$ and $7$ in the first die are equally likely, all possible sums $2,3,\ldots,13$ are equally likely. EDIT: Beaten by above. If the integers must be distinct within a die, it's impossible; easily proven from seeing the sum $2$.

This could have been quite a difficult question back in 1993, if all possibilities were required to be found. If die $A$ has faces labeled $(a_1,a_2,a_3,a_4,a_5,a_6)$, while die $B$ has faces labeled $(b_1,b_2,b_3,b_4,b_5,b_6)$, we can associate to them the exponential generating functions $a(x) = \sum_{k=1}^6 x^{a_k}$ and $b(x) = \sum_{k=1}^6 x^{b_k}$. Then the distribution they need to produce is $f(x) = a(x)b(x)$, and since each of the $12$ numbers must be equally alike obtained ($3$ times, since there are $36$ possibilities), it means we have $f(x) = 3\sum_{k=2}^{13} x^k = 3x^2\sum_{k=0}^{11} x^k = 3x^2\Phi_2(x)\Phi_3(x)\Phi_4(x)\Phi_6(x)\Phi_{12}(x)$. Now we need $a(1)=b(1) = 6$, and also, since $2$ must be obtained, at least one each of the $a_i$'s and $b_j$'s must be equal to $1$. Since $f(1) = 3\cdot 1\cdot \Phi_2(1)\Phi_3(1)\Phi_4(1)\Phi_6(1)\Phi_{12}(1) = 3\cdot 1\cdot 2\cdot 3\cdot 2\cdot 1\cdot 1$, it means we can only have (or obviously reversed) $a(x) = 3x(x+1), b(x) = x(x^2+1)(x^8+x^4+1)$ leading to $(1,1,1,2,2,2)$ and $(1,3,5,7,9,11)$; $a(x) = 3x(x^2+1), b(x) = x(x+1)(x^8+x^4+1)$ leading to $(1,1,1,3,3,3)$ and $(1,2,5,6,9,10)$; $a(x) = 3x(x^3+1), b(x) = x(x^6+1)(x^2+x+1)$ leading to $(1,1,1,4,4,4)$ and $(1,2,3,7,8,9)$; $a(x) = 3x(x^6+1), b(x) = x(x^3+1)(x^2+x+1)$ leading to $(1,1,1,7,7,7)$ and $(1,2,3,4,5,6)$.

mavropnevma: what do you mean by $ \Phi_{2}(x)\Phi_{3}(x)\Phi_{4}(x)\Phi_{6}(x)\Phi_{12}(x) $

Cyclotomic polynomial.