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Recall that we are working modulo $2$. Thus $E(x)$ is a polynomial whose coefficients are $0,1$, and $E(1) \in \{0,1\}$. By definition, $G(x)$ is a factor of $E(x)$ if there exists a polynomial $H(x)$ such that $E(x) = G(x) H(x)$. In this case, we assume that $E(1) = 1$. Since $E(1) = G(1) H(1)$, this forces $G(1) = 1$.