Is it possible to get all possible sums with the same probability if I throw two unfair dice together?

This is a classical problem. Without changing the problem, we can let the digits on the dice be $0, \ldots, 5$ instead of $1, \ldots, 6$ to make our notation easier. Now we make two polynomials: $$ P(x) = \sum_{i=0}^5 p_ix^i,\qquad Q(x) = \sum_{i=0}^5q_ix^i. $$ Now we can succinctly phrase your condition on $p_i, q_i$: it is satisfied if and only if $$ P(x)Q(x) = \frac1{11}\sum_{i=0}^{10} x^i. $$ Let's multiply both sides by $11 \times (x-1)$, and you get $$ 11(x-1)P(x)Q(x) = x^{11} - 1. $$ The 11 zeroes of the polynomial on the right are the 11th roots of unity, which means those are also the zeroes of the polynomial on the left. The term $(x-1)$ takes care of one of the zeroes, and since $P, Q$ are both of degree 5, that means that they each have to have 5 of the other 10 zeroes.

But now note: besides $1$, all of the 11th roots of unity are complex numbers, while $P, Q$ are real polynomials. If a complex number is the root of a real polynomial, then so is its complex conjugate. That means that $P, Q$ must each have an even number of complex zeroes, but we just showed that they also have to have 5 each.

We have reached a contradiction: such $P, Q$, and thus such distributions $p_i, q_i$, do not exist.

Let's assume that this is possible. We can derive a contradiction from this assumption.

The probability of rolling a total of $2$ must be $1/11$, and the probability of rolling a total of $12$ must also be $1/11$, so

$$ \begin{align} p_1 q_1 &= 1/11,\ \text{and}\\ p_6 q_6 &= 1/11. \end{align} $$

The probability of rolling a total of $7$ must also be $1/11$. At the same time, the probability of rolling a total of $7$ is greater than or equal to $p_1 q_6 + p_6 q_1$. So,

$$p_1 q_6 + p_6 q_1 \le 1/11.$$

The numbers $p_1$, $p_6$, $q_1$, and $q_6$ are all probabilities, so they can't be negative. Also, if any one of them were $0$, then either $p_1 q_1$ or $p_6 q_6$ would be $0$, which we already know is not the case. So, all four of these numbers are positive. This means that $p_1 q_6$ and $p_6 q_1$ are both positive, and so

$$ \begin{align} p_1 q_6 &< 1/11,\ \text{and}\\ p_6 q_1 &< 1/11. \end{align} $$

If we combine these inequalities with the equations above, we find that

$$ \begin{align} p_1 q_6 &< p_1 q_1,\\ p_6 q_1 &< p_1 q_1,\\ p_1 q_6 &< p_6 q_6,\ \text{and}\\ p_6 q_1 &< p_6 q_6. \end{align} $$

Since the numbers $p_1$, $p_6$, $q_1$ and $q_6$ are all positive, we may cancel them when they appear on both sides of one of these inequalities. Doing that, we conclude that

$$ \begin{align} q_6 &< q_1,\\ p_6 &< p_1,\\ p_1 &< p_6,\ \text{and}\\ q_1 &< q_6. \end{align} $$

But this is impossible.