jmc

First we note that the four integers are $1,3,5,7$. Then we expand to get $$(abc+abd+acd+bcd)(abcd{)}^{-1}=a^{-1}+b^{-1}+c^{-1}+d^{-1}.$$Finally, we see that (amazingly) each of the four numbers is its own inverse modulo $8$. Thus, $$1^{-1}+3^{-1}+5^{-1}+7^{-1}\equiv 1+3+5+7\equiv 16\equiv \boxed{0}(\mathrm{mod}8).$$