Kilford uses not $E_2$ but something he calls $E_{2, 2}$, which is $$E_{2}(z) - 2E_2(2z) = 1 - 24 \sum_{n \text{ odd}} \sigma_{1}(n) q^n.$$ This is a modular form of weight 2 and level $\Gamma_0(2)$. Similarly $E_{p, 2} = E_2(z) - p E_2(pz)$ is modular of level $\Gamma_0(p)$ for any $p$.
The naive Eisenstein series $$E_{2}(z) = 1 - 24 \sum_{n \in \mathbb{N}} \sigma_{1}(n) q^n$$ is not a modular form of any level.
[EDIT: My original answer contained the following statement, which is obviously wrong: "this follows from the fact that $E_2(-1/z) - z^2 E_2(z)$ is something like $6/\pi \mathrm{Im}(z)$, while for a modular form it would have to be holomorphic."
A hopefully better statement is: we have $E_2(z) - z^{-2} E_2(-1/z) = 2\pi i / z$, and if $E_2$ were modular (of some level) then $z^{-2} E_2(-1/z)$ would also be modular (of some other level) and hence their difference would be modular (for the intersection of the two level groups). But a non-constant rational function cannot be modular of any level.]