The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption.

However, the suggested criterion is not (provably) sufficient in general. I can give two examples. The first is quite pathological, and the second is not, but goes beyond ZFC.

(1) Let $X$ be an uncountable set, and take $\Sigma$ to be the countable-cocountable $\sigma$-algebra (the one generated by singletons). Take $\kappa$ to be the counting measure, *i.e.* a set $S \in \Sigma$ has measure $|S|$ if it is finite and $\infty$ if it is of any infinite cardinality. A function is integrable with respect to this measure iff it is an element of $\ell^1(X)$ (the small $\ell$ is intentional). Therefore a functional in $L^\infty(X,\Sigma,\kappa)$ that is in the image of $L^1(X,\Sigma,\kappa)$ must take a nonzero value on some $\chi_{\{x\}}$ for some $x \in X$.

Now consider the following measure $\nu : \Sigma \rightarrow [0,1]$. It takes the value $0$ on all countable sets and $1$ on all cocountable sets. A quick check shows that it is countably additive, and it is absolutely continuous to $\kappa$. But for all $x \in X$, $\int_X \chi_{\{x\}} \mathrm{d}\nu = 0$, so integration against it does not come from an element of $L^1(X,\Sigma,\kappa)$, even though it satisfies your criterion by the dominated convergence theorem.

(2) The reason why that space is somewhat pathological is that it is not *localizable*, *i.e.* $L^\infty(X,\Sigma,\kappa)$ is not a W$^*$-algebra (isomorphic to a von Neumann algebra). The following is a way of getting a localizable counterexample. Suppose that there exists a real-valued measurable cardinal, *i.e.* that there exists a set $X$ and a countably additive probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons. Such a measure is necessarily not *completely* additive, because it would then be zero (and the set $X$ is necessarily uncountable for the same reason). So there exists a family of sets $(S_i)_{i \in I}$ where $\mu(S_i) = 0$ for all $i \in I$, but $\bigcup_{i \in I}S_i = X$ (for instance, take the family of all singletons).

Now recall the following facts. The spaces $\ell^p(X)$ are exactly the spaces $L^p(X,\mathcal{P}(X),\kappa)$, where $\kappa$ is the counting measure (on $\mathcal{P}(X)$ this time), and each element of $\ell^1(X)$ defines a *completely* additive signed measure $\nu : \mathcal{P}(X) \rightarrow \mathbb{R}$ by integration (because all $\phi \in \ell^1(X)$ have countable support). By the dominated convergence theorem, the element of $L^\infty(X,\mathcal{P},\kappa)^*$ defined by $f \mapsto \int_X f \mathrm{d}\mu$ satisfies your criterion, but cannot be equal to any $f \mapsto \int_X f\phi\mathrm{d}\kappa$ for any $\phi \in \ell^1(X)$ because
$$
\lim_{i \to \infty} \int_X \chi_{S_i} \mathrm{d}\mu = \lim_{i \to \infty} 0 = 0
$$
for all $i \in I$, but for all $\phi \in \ell^1(X)$ that are density functions of probability measures:
$$
\lim_{i \to \infty} \int_X \chi_{S_i} \phi \mathrm{d}\kappa = \int_X \chi_X \phi \mathrm{d}\kappa = 1
$$

**Conclusion**

If you wish to go beyond the $\sigma$-finite case to the localizable case, your criterion is not far off. Essentially all you need to do is replace sequences by nets. The image of $L^1(X,\Sigma,\mu)$ in $L^\infty(X, \Sigma,\mu)^*$ is the *normal* linear functionals, the span of those positive linear functionals that preserve suprema of bounded nets. See Sakai's *C$^*$-algebras and W$^*$-algebras*, Theorem 1.13.2, taking $\mathcal{M} = L^\infty(X,\Sigma,\mu)$ and $\mathcal{M}_* = L^1(X,\Sigma,\mu)$.