I know that it is entirely possible to define on the extended real line $e^{-\infty}=0$. This will work well, because $a0=a e^{-\infty}=e^{\ln a -\infty}=e^{-\infty}$.
But can we introduce such thing as $H$ that has the property of $e^{-H}=e^{a-H}=0$, but at the same time $H+a\ne H\ne b H$?
Yes, of course, this would mean we would not be able to take logarithm of $e^{-H}=0$ (the logarithm would be multivalued) but other than this, does anything get broken?
In other words, given that exponential function of imaginary argument is periodic, why $e^{x-H}$ cannot be a constant (for such special element $H$ and scalar $a$)?
Suppose we add in an a element $H$ to the complex numbers, and we want to have $e^{-H}=0$. We want to still have a integral domain so that $H+a\ne H\ne bH$ for $a\ne0,b\ne1$ and we won't lose multiplicative cancellation. And $H$ must be transcendental over the complex numbers as otherwise polynomials would have too many roots for it to stay an integral domain.
Then we can think about what properties of the exponential function we want to preserve, and see if it's possible. For example, if we want both $e^{x}$ to be always defined, and $e^{a+b}=e^{a}e^{b}$ to always hold, then we reach the contradiction $1=e^{0}=e^{-H}e^{H}=0e^{H}=0$.
It might be tempting to throw out $e^{a+b}=e^{a}e^{b}$, but that follows algebraically from the infinite series definition, as shown in this answer by Kiryl Pesotski. I'm not sure what $e^{x}$ is intended to be/do if we throw out $e^{a+b}=e^{a}e^{b}$, so let's throw out $e^{x}$ always being defined.
As noted, if $a$ is complex, and $e^{a+b}=e^{a}e^{b}$ when everything is defined, then $e^{a-H}=e^{a}e^{-H}=0$. Also note that, for positive integers $n$, we have $e^{-nH}=\left(e^{-H}\right)^{n}=0$. And similarly, if $e^{-H/m}$ is defined for a positive integer $m$, then $\left(e^{-H/m}\right)^{m}=e^{-H}=0$ so that $e^{-H/m}=0$. So with maximal consistent definitions for the exponential function, we have $e^{-rH}=0$ for positive rationals $r$, but $e^{-0H}=e^{0}=1$. This means that the exponential function can't be continuous where defined, or that the function $x\mapsto-Hx$ is not continuous on the nonnegative reals (which has serious consequences - at minimum, I think the ring can't be normed anymore). Either of these would be really unfortunate. Edit: $\exp$ was analytic and single-valued on $\mathbb C$, so losing continuity here is a loss even though I'm, in some sense, using the discontinuity of $0^x$.
If there are other properties you want to preserve, maybe they can be kept, but it seems that, at the very least, we are forced to lose a lot of nice properties.