Can the prime counting function possibly satisfy this functional equation? Let $a,b,n$ be odd natural numbers, $a \gt b$, and $n \gt 1$
Here's a proof by BillyJoe that the function,
$$h(a,b,n) := 2(\lfloor \dfrac{b}{n} \rfloor  - \lfloor\dfrac{b}{2n} \rfloor - \lfloor \dfrac{a-1}{n} \rfloor + \lfloor \dfrac{a - 1}{2n} \rfloor) - \lfloor\dfrac{b}{n} \rfloor + \lfloor \dfrac{b - 1}{n}\rfloor - \lfloor \dfrac{a}{n}\rfloor + \lfloor \dfrac{a-1}{n}\rfloor
$$
counts exactly, the number of pairs of odd numbers $o, o+2 \in [a, b] \subset \Bbb{N}$ such that either $n \mid o$ or $n \mid o + 2$.
Define the inclusion-exclusion operator:
$$
T(f)(a,b,n) := \sum_{u \mid \frac{p_n\#}{2}} (-1)^{\omega(u)}f(a,b,u) 
$$
Then we have that the total number of twin primes lying in the interval $[X^2, X^4]$ for odd $X$ is:
$$
\pi_2(X^2, X^4) = \dfrac{X^4 - X^2}{2} - T(h)(X^2, X^4)
$$
But if there a a finite number of twin primes, then for all $X \geq N_0 \in \Bbb{N}, X$ odd we have:
$$
0 = \pi_2(X^2, X^4) = \dfrac{X^4 - X^2}{2} - T(h)(X^2, X^4, n)
$$
$T$ applied to $\lfloor \dfrac{b}{n} \rfloor - \lfloor\dfrac{b}{2n} \rfloor$ of course equals $\pi(b) - \pi(\sqrt{b}) + 1$ according to algorithms for evaluating $\pi(x)$.  Similarly for $a-1$ in place of $b$.
Thus we have that for all $X \geq N_0$ odd:
$$
0 = \dfrac{X^4 - X^2}{2} - 2 (\pi(X^4) - \pi(X^2) + 1 + \pi(X^2 - 1) - \pi(\sqrt{X^2 - 1}) - 1) + T(\text{the rest})(a,b,n) \\
\iff \\
0 = \dfrac{X^4 - X^2}{2} - 2 (\pi(X^4) - \pi(X^2) + \pi(X^2 - 1) - \pi(\sqrt{X^2 - 1})) + T(\text{the rest})(a,b,n) \\
\iff \\
0 = \dfrac{X^4 - X^2}{2} - 2(\pi(X^4) - \pi(\sqrt{X^2 - 1})) + T(\text{the rest})(a,b,n)
$$
since $\pi(X^2 - 1) = \pi(X^2)$ since $X^2 - 1$ is composite for $X \gt 2$.
And $T(\text{the rest})(a,b,n) = \sum_{u \mid \frac{p_{\pi(X^2)}\#}{2}} (-1)^{\omega(u)} (\lfloor\dfrac{b}{n} \rfloor - \lfloor\dfrac{b - 1}{n}\rfloor - \dots)$.
But that just counts each divisor of $\dfrac{p_{\pi(X^2)}\#}{2}$ that also divides $b$ or in other words:
$$T(\text{the rest}) = \sigma_0(\gcd(\dfrac{p_{\pi(X^2)}\#}{2}, X^4)) - \sigma_0(\gcd(\dfrac{p_{\pi(X^2)}\#}{2}, X^2))
$$

But clearly, because of the way primorial works with a gcd we must have that $T(\text{rest}) = 0$!
Is it possible that under the prime number theorem, $\pi(X)$ can satisfy this functional equation?
$$
\pi(X^4) - \pi(X) = \dfrac{X^4 - X^2}{4}, \forall X \in 2 \Bbb{N} + 1, X \geq N_0
$$
for some $N_0 \in \Bbb{N}$?

We know that $\dfrac{\pi(X^4)}{\pi(X)} \to \dfrac{X^3}{4}$ as $X \to \infty$ viewing $\pi(X)$ as a real function.
How do we reconcile the realness required with the fact the functional equation is only equality necessarily at odd naturals?   Do we use an error term or something? Please demonstrate.

Your question, in short: is there some $N_0 \in \mathbb{N}$ such that
$$
\pi(n^4) - \pi(n) = \dfrac{n^4 - n^2}{4}
$$
for all odd $n$?
The answer is no, for the reason you already gave: it is incompatible with the growth of $\pi(x).$ In particular,
$$
\pi(n^4) - \pi(n) < \dfrac{n^4 - n^2}{4}
$$
for $n\ge4$.
