Sure! Simply add $b+c$ to both sides and note that $2x=0$ mod $2$ for all numbers $x$. So you get $a=a$ mod $2$ implies that $a+2(b+c)= a$ mod $2$. Now subtract $b+c$ on both sides to obtain $a+b+c=a-b-c$ mod $2$. Note that it really didn't matter that we had three numbers instead of $2$ in other words $a+b= a-b$ mod $2$. Another way to see it is the simple rule that addition and subtraction happen to coincide module $2$ ($+=-$ mod $2$). This last rule is often applied in for instance algebraic topology courses when people do not want to care about signs.
For higher numbers $n$ there are of course the very similar identities $a+(n-1)b=a-b$ mod $n$. Altough you won't in general have $a+b=a-b$ mod $n$ if $n>2$.