Why is there no regularity condition in case 1 of the master theorem?
You are asking why the regularity condition is absent in case 1. What you should be asking instead is why the regularity condition is present in case 3. Let's start with the latter question.

Consider the recursion
$$
T(n) = T(n/2) + f(n), \text{ where } f(n)=n^{1 + (\log_2 n \bmod 2)}.
$$
For simplicity I define the function $f$ only for powers of 2, and it is given by $f(2^{2k}) = 2^{2k}$, $f(2^{2k+1}) = 2^{4k+2}$. You can check that $T(n) = \Theta(n^2)$, and so it is not the case that $T(n) = \Theta(f(n))$ (since for $n = 2^k$, $f(n) = n$).

This shows that some regularity condition is indeed necessary in case 3. The proof shows that this particular regularity condition suffices for the conclusion $T(n) = \Theta(f(n))$ to hold.

In case 1, in contrast, the proof works for every function $f(n)$ satisfying the condition $f(n) = O(n^c)$, where $c$ is strictly smaller than the critical exponent. You can try to concoct an example like the one above, but you will necessary fail, since the master theorem states that the condition $f(n) = O(n^c)$ suffices for the conclusion $T(n) = \Theta(n^{c_{crit}})$ to hold.

You can still ask whether there is some deeper reason that the regularity condition is needed in case 3 but not in case 1. I think that the example above is very suggestive: the reason is that in case 3 the answer involves $f(n)$, whereas in case 1 it doesn't. Therefore if $f(n)$ is not well-behaved then it could show up in case 3, but is irrelevant for case 1.