Substituting $n=1$ into the equation $a_{n+1}=a_{n}+a_{n+2}-1$ gives $a_{2}=a_{1}+a_{3}-1$. Since $a_{1}=x$ and $a_{3}=y$, then $a_{2}=x+y-1$. Rearranging the given equation, we obtain $a_{n+2}=a_{n+1}-a_{n}+1$ for each $n \geq 1$. Thus, $a_{4}=a_{3}-a_{2}+1=y-(x+y-1)+1=2-x$, $a_{5}=a_{4}-a_{3}+1=(2-x)-y+1=3-x-y$, $a_{6}=a_{5}-a_{4}+1=(3-x-y)-(2-x)+1=2-y$, $a_{7}=a_{6}-a_{5}+1=(2-y)-(3-x-y)+1=x$, $a_{8}=a_{7}-a_{6}+1=x-(2-y)+1=x+y-1$. Since $a_{7}=a_{1}$ and $a_{8}=a_{2}$ and each term in the sequence depends only on the previous two terms, then the sequence repeats each 6 terms. Now $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=x+(x+y-1)+y+(2-x)+(3-x-y)+(2-y)=6$ which means that the sum of each successive group of 6 terms is also equal to 6. We note that $2016=6 \cdot 336$ and so the 2016th term is the end of a group of 6 terms, which means that the sum of the first 2016 terms in the sequence is $6 \cdot 336=2016$. Finally, $a_{2017}=a_{1}=x$ and $a_{2018}=a_{2}=x+y-1$. Thus, the sum of the first 2018 terms is $2016+x+(x+y-1)=2x+y+2015$.