I claim that the first player (call them $A$) has a winning strategy over the second player (call them $B$).
First, $A$ takes $1$ coin so the total reduces to $2018$. Now if $B$ takes $x$ coins, $A$ responds by taking $101-x$ coins. Since $2 \leq x \leq 100$ then $1 \leq 101-x \leq 99$ and the parity matches, so this is always valid. In this way, $A$ can force the total number of coins to reduce by exactly $101$ each "round".
After $19$ such rounds, there will be $2018-(19*101) = 99$ coins left, with $B$ to move. $B$ can only take an even number of coins, so they can take between $2$ and $98$ coins. No matter what they do, there will be at least one coin left, and it will be an odd number less than $100$. So $A$ ends the game by taking all remaining coins and $B$ can no longer make a move. $\blacksquare$