Can $y=10^{-x}$ be converted into an equivalent $y=\mathrm{e}^{-kx}$?

Yes, it can be. Notice that

$${y=10^{-x}\geq 0}$$

Hence ${\log(10^{-x})}$ is well defined, and so

$${10^{-x}=e^{\log(10^{-x})}=e^{-x\log(10)}=e^{-\log(10)x}}$$

And so

$${k=\log(10)\approx 2.30258509....}$$

Hint. $10=e^{\ln 10} \phantom{stuff}$

$ y = 10^{-x} = e^{\log(10^{-x})} = e^{-x \log(10)} $ so $k = \log(10) = 2.3025...$