I further show that this sum distribution is impossible for any pair of dice, no matter how many faces they have, what numbers are on those faces, and what weights each face has, barring the trivial case where one die always rolls the same result.
Translate the dice weightings into generating functions
$$A(x) = \sum a_i x^i$$
and likewise for $B(x)$. Since dice sum is distribution convolution is generating function product, we're required to have
$$A(x) B(x) = \frac{x^2+x^3+\dots + x^{11} + x^{12}}{11}$$
Express the geometric series on the RHS as
$$A(x) B(x) = \frac{x^2}{11} \cdot \frac{1-x^{11}}{1-x}$$
The roots of $1-x^{11}$ are the powers of the primitive elevent roots of unity $\omega = e^{2\pi i/11}$, which are $1, \omega, \omega^2, \dots, \omega^{11}$. The $1-x$ in the denominator removes the root $x=1$. So, we can factor into linear terms
$$A(x) B(x) = \frac{x^2}{11} \cdot \prod_{j=1}^{10} (x-\omega^j)$$
Now, because of unique factorization, these linear terms must be split across $A$ and $B$. Moreover, since the coefficients of $A$ and $B$ are probabilities, the resulting polynomials must have non-negative real coefficients.
At this point, we're down to a finite number $2^{10}$ of possibilities, so we could try all of them by computer and check that none work. But, we can simplify the search by hand first.
First, a real polynomial must equal its conjugate, so its complex roots must come in conjugate pairs. Therefore, the conjugate roots must "stick" together when being split into $A$ and $B$. So, we can group them into quadratic units
$$(x-\omega^j)(x-\overline{\omega^j}) = x^2 - 2\mathrm{Re}(\omega^j)x + 1$$
for $j=1,2,3,4,5$.
Now, note that for a product of quadratic units, the coefficient of $x$ is the sum of all $-2\mathrm{Re}(\omega^j)$ of the $j$'s included. Since this coefficient must be positive, the sum of the real parts of the included roots of unity must be negative.
These real parts are roughly $0.84, 0.42, -0.14, -0.65, -0.96$. Already we see that $0.84$ must be grouped with $-0.96$ to make a negative sum, since the other negatives aren't enough. Then, $0.42$ must be grouped with $-0.65$, since the $-0.14$ and the $-0.12$ from the previous pairing aren't enough to make it negative.
So, we have three groups $\{\omega^1,\omega^5\},\{\omega^2,\omega^4\},\{\omega^3\}$ split into two sides. Putting them all onto one side would result in a trivial monomial on the other side, giving a die that always rolls the same. So, we must have one group alone and the other two groups together. Checking all three possibilities directly, each one gives a polynomial that contains negative coefficients. So, the desired split is not possible, and no solution exists.