The answer is yes. We will show that a pair of such sets exists. Rewrite the desired inequality as
$$\left| \sum_{a \in A} (1-y)^a - \sum_{b \in B} (1-y)^b \right| < y^{2011} \quad (34)$$
for every $y$ such that $0 < y < 1$.

Step 1: First, we show that there exist two different sets $A'$, $B'$ of $2011^2$ positive integers each such that for $k = 0, 1, 2, \dots, 2011$, we have
$$\sum_{a \in A'} \binom{a}{k} = \sum_{b \in B'} \binom{b}{k}. \quad (35)$$
Let $N$ be a large positive integer whose value will be determined later. For any subset $S$ of $\{1, 2, \dots, N\}$ having $2011^2$ elements, form the 2011-tuple
$$c(S) = \left( \sum_{s \in S} \binom{s}{1}, \sum_{s \in S} \binom{s}{2}, \dots, \sum_{s \in S} \binom{s}{2011} \right).$$
The first element of $c(S)$ is an integer between $1$ and $2011^2 N$; the second lies between $1$ and $2011^2 N^2$, and so forth. Thus there are at most
$$2011^{4022} \cdot N^{1+2+\dots+2011} = 2011^{4022} \cdot N^{\binom{2012}{2}}$$
possible values that $c(S)$ can take on. On the other hand, the number of possible sets $S$ is $\binom{N}{2011^2}$, which is greater than $2011^{4022} \cdot N^{\binom{2012}{2}}$ when $N$ is sufficiently large. Hence, by the pigeonhole principle, some two subsets $A'$ and $B'$ have $c(A') = c(B')$. These are our sets $A'$ and $B'$ satisfying (35).

Step 2: Now consider expanding the binomials in the expression
$$\sum_{a \in A'} (1-y)^a - \sum_{b \in B'} (1-y)^b.$$
By (35), all the terms of degree less than or equal to $2011$ cancel, meaning that all remaining terms have degree $2012$ or higher. Let $M$ be the sum of the absolute values of the coefficients of all the remaining terms. We thus have
$$\left| \sum_{a \in A'} (1-y)^a - \sum_{b \in B'} (1-y)^b \right| \leq M y^{2012} \qquad (36)$$
for all $y$ with $0 < y < 1$.

Step 3: Finally, we claim that
$$(1 - y)^M < \frac{1}{My} \qquad (37)$$
for all $y$. For example, this follows from the AM-GM inequality, via
$$My(1-y)^M \leq \left( \frac{My + \overbrace{(1-y) + \dots + (1-y)}^{M}}{M+1} \right)^{M+1} = \left( \frac{M}{M+1} \right)^{M+1} < 1.$$

Step 4: To finish, let $A = \{a + M \mid a \in A'\}$ and $B = \{b + M \mid b \in B'\}$. Multiplying (36) and (37) gives (34) for this choice of $A$ and $B$, as required.