Let $E$ be a set, $\mathcal E\subseteq2^E$ with $\emptyset\in\mathcal E$ and $\eta:\mathcal E\to\mathbb R$ with $\eta(\emptyset)=0$. If $B\subseteq E$, let $$|\eta|(B):=\sup\sum_{i=1}^k|\eta(B_i)|,$$ where the supremum is taken over all $k\in\mathbb N$ and disjoint $B_1,\ldots,B_k\in\mathcal E$ with $\bigcup_{i=1}^kB_i\subseteq B$.
I want to show that $$|\eta|(E)=\sup_{B\in\mathcal E}\left(\eta(B)-\eta(B^c)\right)=\sup_{B\in\mathcal E)}\left(|\eta(B)|+|\eta(B^c)|\right)\tag1.$$
If necessary, impose further assumptions on $\mathcal E$ (e.g. $B_1\cup B_2\in\mathcal E$ for all $B_1,B_2\in\mathcal E$) or $\eta$ (e.g. additivity).
I guess we can somehow argue by considering $B=\bigcup_{i:\eta(B_i)\ge0}B_i$.
Best Answer
In this answer I'll assume that $\mathcal{E}$ is closed under finite unions and complements and that $\eta$ is finitely additive. Note that you really use that $\mathcal{E}$ is closed under complements for the result you want to make sense so these are only the additional assumptions formulated in your question.
Then the result is trivial from your suggestion. Indeed, for disjoint $B_1, \dots, B_k \in \mathcal{E}$, let $B$ be defined as in the question. Then $$\sum_{i=1}^k |\eta(B_i)| = \eta(B) - \eta(B^c) = |\eta(B)| + |\eta(B^c)|$$ by finite additivity. Now taking the $\sup$ in each of these terms gives the result.
It's also obvious that both finite additivity of $\eta$ and $\mathcal{E}$ being closed under unions are necessary.
If you don't assume that $\mathcal{E}$ is closed under finite unions then consider $E = \{0,1,2,3\}$, $\mathcal{E}$ to be the set containing $\emptyset, E$, all singletons in $E$ and all three element subsets of $E$ and set $\eta = \delta_0 + \delta_1 - \delta_2 - \delta_3$. Then it is easy to check that $|\eta|(E) = 4$ but $\sup_{B \in \mathcal{E}} |\eta(B)| + |\eta(B^c)| = 2$. Note that this $\eta$ is even the restriction of a signed measure to $\mathcal{E}$.
If you assume that $\mathcal{E}$ is closed under finite unions but $\eta$ is not additive things also go wrong. For example, let $E = \{0,1,2,3\}$, $\mathcal{E} = \mathcal{P}(E)$ and set $\eta(\emptyset) = 0$ and $\eta(A) = 1$ for all other $A \in \mathcal{E}$. Again we have that $|\eta|(E) = 4$ and $\sup_{B \in \mathcal{E}} |\eta(B)| + |\eta(B^c)| = 2$.