Looking at the current version of your post, we have
$(1-e^{-x})e^{-y}=e^{-y}-e^{-x}e^{-y}>-e^{-x}e^{-y},$ since $e^t$ is positive for all real $t$. However, we can't take the logarithm of the right-hand side. It's negative.
Update:
The old version was $(1-e^{-x_1})(1-e^{-x_2})e^{-x_3}=e^{-x_3}-e^{-x_1-x_3}-e^{-x_2-x_3}+e^{-x_1-x_2-x_3},$ and you wanted to know if that was greater than or equal to $(-e^{-x_1})(-e^{-x_2})e^{-x_3}=e^{-x_1-x_2-x_3}$ for all positive $x_1,x_2,x_3$. Note, then, that the following are equivalent (bearing in mind the positivity of $e^t$):
$(1-e^{-x_1})(1-e^{-x_2})e^{-x_3}\geq e^{-x_1-x_2-x_3}$
$e^{-x_3}-e^{-x_1-x_3}-e^{-x_2-x_3}\geq 0$
$e^{-x_3}(1-e^{-x_1}-e^{-x_2})\geq 0$
$1-e^{-x_1}-e^{-x_2}\geq 0$
This need not hold. In fact, for any $x_2>0$, there is some $x_1>0$ such that the inequality fails to hold. (Let me know if you're interested in a proof of that fact.)