Why do people differentiate between discrete and continuous decay?
If a block of whatever is decaying to half it’s amount ever $x$ years, it is perfectly natural, and continuous to say: $m_0(1/2)^{t/x}=m(t)$ this is a continuous function, not discrete, and it exactly models the decay of such an object. It does not make sense why simply rewriting the same function with base $e$ makes it in some way more continous. Perhaps this can be seen even more clearly as follows:
$$\frac{dm}{dt}=km \implies dt=\frac{dm}{km}=\frac{dm \cdot \ln(1/2)}{km \cdot \ln(1/2)}$$
Integrating,
$$\int{dt}=\frac{1}{k \ln(1/2)} \int{\frac{\ln(1/2)}{m}dm} \implies \left(\frac12 \right)^{\frac{t}{k \ln(1/2)}} \cdot \left (\frac12 \right)^c=m(t)$$
No, changing bases from, let's say $2$, to $e$ does not make it more "continuous".
Rather, this is convention, as $e$ is the base of the natural logarithm and much easier to use in calculus.
In addition, converting different bases to base-e allows to easily compare growth/decay of different functions.