Note that we are not interested in differentiating the gate $U(\theta)$, but some expectation-value-based function $E(\theta)$, which "contains the gate twice", if you will: $$E(\theta)=\langle \psi_0|U(\theta)^\dagger H U(\theta)|\psi_0\rangle$$
This function can be shown to be a Fourier series, with a single frequency for gates that satisfy the original shift rule. This means that
$$E(\theta) = a_0 + a_1\cos(\theta)+ b_1\sin(\theta)$$
where I assumed the single frequency to be $1$ for simplicity.
Furthermore, without loss of generality, we may assume that we want the derivative of $E$ at $0$, because we may absorb the "rest" of $U(\theta)$ in the state $\psi_0$.
This means, we want to find $E'(0)=b_1$. At the same time, $E(0)=a_0+a_1$, i.e. the un-shifted evaluation does not contain any information about $b_1$, and does not allow us to resolve $a_0$ and $a_1$, but only their sum.
The conclusion from this is, that we need two evaluations that are somehow shifted away from $0$ (or a multiple of $\pi$ for that matter, because the $\sin(\theta)$ part vanishes at those points).
You might be happy to hear, that computing the second-order derivative $E''(0)$ is indeed possible with just one additional execution:
$$E''(0) = a_1 = (E(0) - E(\pi)) / 2$$
This last statement appears for example in Mari, Bromley, Killoran (2021).
For more information, you could be interested in
