Tried to prove that $e$ is irrational but ended up proving it's not a real number.

You went wrong in the last step. As you're fully aware, $e^{2i \pi} = 1$, just as $e^0 = 1$. This means that the complex exponential function is no longer injective, literally meaning that $e^z = e^w$ no longer implies $z = w$, with $z = 2 i \pi$ and $w = 0$ being a counterexample.

So, what does $e^z = e^w$ actually imply? We have $$e^z = e^w \implies \frac{e^z}{e^w} = 1 \implies e^{z - w} = 1.$$

If we write $z - w = x + iy$, where $x, y \in \Bbb{R}$, this yields $$1 = e^{x + iy} = e^x(\cos(y) + i \sin(y)),$$ which implies $e^x = 1$ from taking the modulus of both sides, and $y = 2\pi k$ for some integer $k$. Since $x \in \Bbb{R}$, we must have $x = 0$, so $$z - w = 2 i \pi k$$ for some $k \in \Bbb{Z}$.

So, given $p^{2 i \pi} = q^{2 i \pi}$, we can conclude, by definition of complex exponentiation, $$e^{2 i \pi \log p} = e^{2 i \pi \log q}$$ and thus, for some $k \in \Bbb{Z}$, $$2 i \pi \log p = 2 i \pi \log q + 2 i \pi k \implies \log(p / q) = k \implies p / q = e^k.$$ In this case, $k = 1$; there is no contradiction.