We will show that the second player can win regardless of the first player's actions. Obviously, none of the players will write $0$ in any step.

Notice that $10^r \equiv (-1)^r \pmod{11}$, so the following criterion for divisibility by $11$ holds:
$$\overline{a_k a_{k+1} \dots a_n} \equiv a_n - a_{n-1} + a_{n-2} - \dots + (-1)^{n-k} a_k \pmod{11}$$
Denote by $N_k$ the remainder when dividing $\overline{a_k a_{k+1} \dots a_n}$ by $11$, for $k = 1, \dots, n$. If, after the $n^{th}$ move, the numbers $N_1, \dots, N_n$ are pairwise distinct, then by the mentioned criterion, in the next move we obtain the numbers $a_{n+1} - N_1, \dots, a_{n+1} - N_n, a_{n+1} \pmod{11}$ which are also pairwise distinct since $a_{n+1} \neq 0$. Inductively, we conclude that after each step of the game (for $n \le 10$), the numbers $N_1, \dots, N_n$ are distinct. Because of the way in which $N_1, \dots, N_n$ transform at each move, we conclude that there are at most $n+1$ digits which, if the player writes them, will cause him to lose.

Assume that the game lasts for at least nine moves. The second player loses if and only if, after the ninth move, the set $\{N_1, N_2, \dots, N_9\}$ is $\{1, 2, \dots, 9\}$, i.e. the second player wins if and only if $10$ is among the numbers $N_1, N_2, \dots, N_9$.

If, after the eighth move, $\{N_1, ..., N_8\}$ lacks two numbers between $1$ and $10$ which are not consecutive, then, no matter what the first player chooses in the ninth move, one of the numbers $N_1, ..., N_9$ must be $10$. Namely, if the first player chooses $X$, then there is $k$ such that $N_k = X + 1$ (in eighth move). Hence, after the ninth move we obtain
$$N_k \equiv X - (X + 1) \equiv -1 \equiv 10 \pmod{11}.$$
Let us show that the second player can ensure that after the eighth move, among the numbers $N_1, ..., N_8$ there are no two consecutive numbers. Certainly, the second player can ensure that the game lasts at least seven moves. Let
$$\{N_1, ..., N_7\} = \{1, 2, ..., 10\} \setminus \{X, Y, Z\}.$$
If among the numbers $X$, $Y$ and $Z$ there are no consecutive ones, then the second player can choose any of them. If $Y = X + 1$, then the second player can write one of the numbers $X$ or $X + 1$ so that, after the eighth move, there are two consecutive numbers missing from $N_1, ..., N_8$. Namely, if $Z = X - 1$, the second player writes $X$, and if $Z \neq X - 1$, the second player writes $X + 1$.