It turns out that the equivalence does not hold for finitely additive measures.
The idea for this answer is from https://mathoverflow.net/questions/155836/additive-measure-on-countable-algebras.
First observe that if $\mathcal{F}$ is an ultra filter on a set $X$, then we can define a finitely additive measure $\mu$ on $\rm{Pot}(X)$ by setting $\mu(A) = 1$ for $A \in \mathcal{F}$ and $\mu(A) = 0$ for $A \notin \mathcal{F}$.
To see this, let $A,B \subset X$ with $A \cap B = \emptyset$. There are four cases:
- $A \in \mathcal{F}$, $B \notin \mathcal{F}$. Then $A \cup B \in \mathcal{F}$ because of $A \cup B \supset A$. Hence $\mu(A) + \mu(B) = 1 = \mu(A \cup B)$.
- $A \notin \mathcal{F}$, $B \in \mathcal{F}$. Interchange $A,B$ and apply (1).
- $A,B \in \mathcal{F}$. This would imply $\emptyset = A \cap B \in \mathcal{F}$, in contradiction to the definition of a filter.
- $A ,B \notin \mathcal{F}$. As $\mathcal{F}$ is an ultra filter, this yields $A^c,B^c \in \mathcal{F}$ and thus $(A \cup B)^c =A^c \cap B^c \in \mathcal{F}$, which implies $A \cup B \notin \mathcal{F}$ and hence $\mu(A) + \mu(B) = 0= \mu(A \cup B)$.
Now note that
$$
\mathcal{N} := \{M \subset [0,1] \,\mid\, M^c \text{ is a Lebesgue null-set}\}
$$
is a filter. To see this, note that if $M \subset M'$ with $M \in \mathcal{N}$, then $(M')^c \subset M^c$ is a null-set. Furthermore, if $M,M' \in \mathcal{N}$, then $(M \cap M')^c = M^c \cup (M')^c$ is also a null-set. Finally, $\emptyset^c = [0,1]$ is not a null-set.
Using Zorn's Lemma, we see that there is an ultra-filter $\mathcal{F} \supset \mathcal{N}$. Define the measure $\mu$ as above.
Set
$$
\nu : \mathcal{L} \to [0,\infty), A \mapsto \lambda(A) + 2\mu(A),
$$
where $\lambda$ is Lebesgue-measure on $[0,1]$ and $\mathcal{L}$ is the Lebesgue $\sigma$-algebra on $[0,1]$. Note that $\nu$ is finitely additive.
Observe that $\nu([0,1]) = 3$, but for any $A \in \mathcal{L}$, we have
$$
\nu\left(A\right)=\begin{cases}
\lambda\left(A\right)\leq1, & \mu\left(A\right)=0,\\
\lambda\left(A\right)+2\mu\left(A\right)\geq2, & \mu\left(A\right)=1.
\end{cases}
$$
Hence, $\nu$ does not assume any values in $(1,2)$, so that $\nu$ does not satisfy Definition 2.
But $\nu$ satisfies Definition 1, because for $A \in \mathcal{L}$ with $\nu(A) > 0$, we must have $\lambda(A) > 0$ (otherwise $A^c \in \mathcal{N} \subset \mathcal{F}$, which implies $A \notin \mathcal{F}$ and hence $\mu(A) = 0$ which implies $\nu(A) = 0$). As $\lambda$ is atomless, there is some $B \subset A$ with $0 < \lambda(B) < \lambda(A)$, which yields
$$
0 < \lambda(B) \leq \nu(B) = \lambda(B) + 2\mu(B) < \lambda(A) + 2\mu(B) \leq \nu(A).
$$
