Solution 1:

If you want the properties $$e^{x+y}=e^xe^y\quad\hbox{for all $x,y$}$$ and $$e^0=1$$ to remain true, then we have $$e^xe^{-x}=1\quad\hbox{for all $x$}$$ and so $e^x$ can never be zero.

Solution 2:

Adding to what David said, even if you give up on the homomorphism part of the exponential, the series definition (which holds in any Banach algebra, which includes the number systems) make so that $e^X e^{-X}=1$, so $e^X$ can't be $0$.