Mikael de la Salle points out this is true when $E$ is separable, as shown in Corollary V.12.8 of Conway, A Course in Functional Analysis, 2e.
For a non-separable counterexample, consider the uncountable ordinal space $[0, \omega_1]$, which is compact Hausdorff, and $E = C([0, \omega_1])$. By the Riesz representation theorem, $E'$ is the space of signed Radon measures $\mu$ on $[0, \omega_1]$ with its total variation norm. Let $\varphi(\mu) = \mu(\{\omega_1\})$. This is clearly not represented by any vector in $E$ since the function $1_{\{\omega_1\}}$ is not continuous, but I claim $\varphi$ is sequentially $\sigma(E', E)$ continuous.
Let $\mu_n$ be a sequence converging to 0 in $\sigma(E', E)$ and fix $\epsilon > 0$. Since each $\mu_n$ is Radon, so is its total variation measure $|\mu_n|$, and thus we can approximate $\{\omega_1\}$ in $|\mu_n|$-measure from outside by open sets. So there exists $\alpha_n < \omega_1$ such that $|\mu_n|((\alpha_n, \omega_1)) < \epsilon$. Let $\alpha = \sup_n \alpha_n < \omega_1$; then $|\mu_n((\alpha, \omega_1))| \le |\mu_n|((\alpha, \omega_1)) < \epsilon$ for every $n$.
Define $f : [0, \omega_1] \to \mathbb{R}$ by $$f(x) = \begin{cases} 0, & x \le \alpha \\ 1, & x > \alpha \end{cases}$$ and note that $f$ is continuous. Now $$\varphi(\mu_n) = \mu_n(\{\omega_1\}) = \mu_n((\alpha, \omega_1]) - \mu_n((\alpha, \omega_1)) = \int f\,d\mu_n - \mu_n((\alpha, \omega_1)).$$
But by assumption $\int f\,d\mu_n \to 0$, and $|\mu_n((\alpha, \omega_1))| < \epsilon$, so we conclude $\varphi(\mu_n) \to 0$.