Answer: Bob wins.

It is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \div 11$, where $S_o, S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.

Let Ann use the numbers $a$, $c$, and $y$ for her first, second, and third moves respectively; let Bob use the numbers $b$ and $x$ for his first and second moves respectively.

The following cases are possible after two Ann's moves and one Bob's move:

1. $a$ and $c$ occupy two odd positions, $b$ occupies the even position of the asterisks;
2. $a$ and $c$ occupy two even positions, $b$ occupies the odd position of the asterisks (different from the first position);
3. $a$ and $c$ occupy one even and one odd positions, $b$ occupies the even position of the asterisks.

Let $X = \{x_1, x_2, \dots, x_7\}$ be the set of unused digits before the second Bob's move.

In the first case Bob replaces the asterisk on the even position by $x$. Then $S_o = a + c + y$, $S_e = b + x$. Let $r = b - a - c$. Now the congruence $S_o \equiv S_e \pmod{11}$ is equivalent to the congruence $y \equiv x + r \pmod{11}$.

If $r \equiv 0 \pmod{11}$, then $B$ can use any digit from the set $X$ as the number $x$ because the congruence $y \equiv x \pmod{11}$ is impossible if $x$ and $y$ are the distinct digits.

Let $r \not\equiv 0 \pmod{11}$. Suppose that for any digit $x_i \in X$ there exists a digit $y_i \ne x_i$, $y_i \in X$ such that $y_i \equiv x_i + r \pmod{11}$, $i = 1, 2, \dots, 7$. Summing these seven congruences we obtain
$$y_1 + y_2 + \dots + y_7 \equiv x_1 + x_2 + \dots + x_7 + 7r \pmod{11}.$$
Since it is evident that $y_1 + y_2 + \dots + y_7 = x_1 + x_2 + \dots + x_7$, we see that the obtained congruence is equivalent to the congruence $7r \equiv 0 \pmod{11}$, which is impossible since $r \not\equiv 0 \pmod{11}$. Therefore, there exists a number $x \in X$ such that $y \not\equiv x + r \pmod{11}$ for any digit $y \in X$ different from $x$. So Bob can use this number $x$ for his second move to win.

Consider two other cases. In these cases Bob replaces the asterisk on the odd position (different from the first one) by $x$. Then we have
$$S_o - S_e = (b + x + y) - (a + c) \quad (1)$$
for the second case, and
$$S_o - S_e = (c + x + y) - (a + b) \quad (2)$$
or
$$S_o - S_e = (a + x + y) - (c + b) \quad (3)$$
for the third one.

Set $r = a + c - b$ in (1), $r = a + b - c$ in (2), and $r = b + c - a$ in (3).

Therefore, the congruence $S_o \equiv S_e \pmod{11}$ is equivalent to the congruence $x+y \equiv r \pmod{11}$ for all cases. Suppose that for any digit $x_i \in X$ there exists a digit $y_i \in X$ such that $x_i + y_i \equiv r \pmod{11}$, $y_i \neq x_i$, $i = 1, 2, \dots, 7$. It follows that all digits from $X$ are partitioned into the pairs of distinct digits with the sum which is congruent to $r$ modulo $11$, but it is impossible since the number of digits in $X$ is odd. Therefore, there exists a number $x \in X$ such that $y+x \not\equiv r \pmod{11}$ for any digit $y \in X$ different from $x$. So Bob can use this number $x$ for his second move to win.