Answer: Player 1 has a winning strategy for all initial values, including $a_0 = 2^{2025}$.
Let us consider small values of $a_0$.

Case 1: $a_0 \le 8$. To avoid repeating a number three times, each player must choose $a_n = a_{n-1} + 1$. Then player 1 can increment three times in a row:
$$a_0, a_0 + 1, a_0 + 2, a_0 + 3,$$
which forms an arithmetic progression of length 4, ending the game with a win for player 1.

Case 2: $a_0 = 9$. If player 1 plays $a_1 = 10$, then:
- If player 2 plays $a_2 = 11$, player 1 responds with $a_3 = 12$ and wins via arithmetic progression.
- If player 2 instead plays $a_2 = S(10) = 1$, the position reduces to the previously analyzed case $a_0 = 1$, where player 1 also wins.

Case 3: $a_0 \ge 10$. We claim that using the digit sum operation $S(a)$ leads to losing positions. Note that for $a \ge 10$, we always have
$$S(a) \le a - 9.$$
Hence, each application of $S$ strictly decreases the value by at least 9, eventually reducing to a single-digit number, which as previously shown results in a loss for the player making that move.
Therefore, under optimal play, both players avoid using $S$ and only apply $a_n = a_{n-1} + 1$. Then, player 1 forces: $a_0, a_0 + 1, a_0 + 2, a_0 + 3$, forming an arithmetic progression of length 4, thereby winning.