My calculator gives a solution to $e^x=0$, this on itself quite strange, but the value is also interesting: $x=-9,9997457056636 \times 10^{98}$. My best guess is that it has to do with the limit of $x$ approaches minus infinity, but then why not choose $1 \times 10^{99}$ and not a very specific number?
What perhaps also helps is a picture of the result and the make of the calculator, it is a TI-84 Plus CE-T
First, it should be noted that the TI-84 Plus does not have a built-in computer algebra system like the TI-89 or TI-92. It can only give numerical answers.
Also note that, according to the manual, “Values are stored in memory using up to 14 digits with a two-digit exponent.” Thus, the greatest (absolute value) number that can be represented on this calculator is $9.9999999999999 \times 10^{99}$. Conversely, the smallest positive number that's distinguishable from zero is $1.0 \times 10^{-99}$. Numbers are stored in a BCD-based format so that exact decimal fractions can be represented exactly. Contrast with the double-precision binary floating-point popular on personal computers, which have a wider range (up to $1.7976931348623157 \times 10^{308}$), but cannot exactly represent $0.1$ or $0.01$.
So, according to the rules of TI-84+ arithmetic, any $x < \ln(10^{-99}) \approx -227.95592420641$ is a valid solution to the equation $e^x = 0$, since the calculation of $e^x$ will underflow.
Still, there are a lot of representable numbers between $-9.9999999999999 \times 10^{99}$ and $-227.95592420642$, so why pick $−9.9997457056636 \times 10^{98}$ in particular, rather than a nice round $-10^{99}$? Well, that's hard to answer without seeing the exact code for the root-finding algorithm. (I tried to speculate, but couldn't get the math to work out.)
Point is, numerical solvers behave weirdly when a function has an asymptote, so be skeptical when your calculator shows you an unusually large or small number.