Cool problem! Let $a,b$ be the game numbers of Alice and Bob respectively, $c$ be the starting number chosen by Bob and $x,y$ be the numbers of turns Alice and Bob have respectively until they reach or exceed $2019$. It is easy to see that $a,b\in \{19,20\}$, $c\in\{9,10\}$ and $x-y\in\{0,1\}$. Now also let $x=y+d$ with $d\in\{0,1\}$. Now putting the equation together we get, $ax+by+c=2019\Longleftrightarrow$ $\Longleftrightarrow ay+ad+by+c=2019\Longleftrightarrow$ $\Longleftrightarrow (a+b)y+ad+c=2019\Longleftrightarrow$ $\Longleftrightarrow 39y+ad+c=2019$. Now because $\left\lfloor\tfrac {2019}{39}\right\rfloor=51\Rightarrow y=51$ and now $1989+ad+c=2019$. So, $30=ad+c\leq 20\cdot 1+10=30$. This means that, $c=10,a=20,d=1,b=19,y=51,x=52$. This gives that if Alice picks $20$ as her game number and Bob picks $10$ as the starting number, Alice wins. Otherwise the game always ends in a tie. So we conclude that, $\bullet$ Bob can never win. $\bullet$ Bob can just choose $9$ as the starting number to prevent Alice from winning.