First of all, there's no reason to restrict the definition of finite additivity to finite spaces. It's just that if a finitely additive measure is defined on a finite space, then, trivially, it's countably additive.
We can extend finite additivity as you've defined it by induction. Your axiom implies that, for any $n \in \mathbb{N}$, if $A_1,...,A_n$ is a sequence of pairwise disjoint events, then $$\sum_nP(A_n) = P(\cup_n A_n).$$
But finite additivity does not imply countable additivity. Indeed, consider the "fair integer lottery" (the uniform distribution over integers) that was of interest to De Finetti. This is a finitely additive probability measure defined on all subsets of $\mathbb{Z}$ such that $P(\{ z\})=0$ for all $z \in \mathbb{Z}$. Countable additivity fails because
$$P(\mathbb{Z}) = 1 \neq 0 = \sum_z P(\{z \}).$$
Note that this is analogous to Lebesgue measure on $[0,1]$, for which we have countable additivity but not uncountable additivity.
Now, it's a little bit difficult to show rigorously that the measure described above actually exists. A relatively easy way to do it uses an ultrafilter. The existence of the required ultrafilter is usually established by using the axiom of choice or its equivalent, Zorn's lemma. (I wrote a little bit about this here.)
In fact, any ultrafilter $\mathcal{U}$ defines a finitely additive measure by setting $P(U)=1$ if $U \in \mathcal{U}$ and $P(U) = 0$ if $U \notin \mathcal{U}$. See this for example.
On the other hand, it has been shown that any purely finitely additive measure is non-constructible: the existence of such a measure cannot be proved with the ZF axioms of set theory alone. See this paper.
Here is a counterexample. Pick a nonprincipal ultrafilter $U$ on $\mathbb{N}$ and consider the finitely additive probability space $(\mathbb{N},\mathcal{P}(\mathbb{N}),P)$ where $$P(A)=\sum_{n\in A}\frac{1}{2^{n+2}}$$ if $A\not\in U$ and $$P(A)=\frac{1}{2}+\sum_{n\in A}\frac{1}{2^{n+2}}$$ if $A\in U$. (So, we have a weighted counting measure on $\mathbb{N}$ with total weight $\frac{1}{2}$, and we give an extra $\frac{1}{2}$ weight to being in $U$.) This is not countably additive since $U$ is nonprincipal so the measures of all the singletons only add up to $\frac{1}{2}$. However, I claim it satisfies the Borel-Cantelli lemma.
Indeed suppose a sequence of sets $(A_n)$ satisfies $\sum P(A_n)<\infty$. If some $k$ were in infinitely many $A_n$, then $P(A_n)$ would be at least $\frac{1}{2^{k+2}}$ for infinitely many $n$, and $\sum P(A_n)$ would diverge. Thus no $k$ is in infinitely many $A_n$, and $\limsup A_n=\emptyset$, so $P(\limsup A_n)=0$.
Best Answer
If $P(A \cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then
for $A_1 \ldots A_n$ disjoints sets, one can consider $P(A_1 \cup A_2 \cup \ldots \cup A_n) = P(A_1 \cup B) $ where $B = A_2 \cup \ldots \cup A_n$ is disjoint from $A$.
From the first property: $P(A_1 \cup B) = P(A_1) + P(B)$
Proceed in the same manner $n$ times (or apply induction) to obtain that $P(A_1 \cup A_2 \cup \ldots \cup A_n) = P(A_1) + P( A_2 ) + \ldots P (A_n)$
And that is finite aditivity.
Finite additivity does not imply countable aditivity. For instance consider on $(\mathbb{N} , \mathcal{P}(\mathbb{N})\,)$ be such that $$P(A) = \begin{cases}0 & \text{ if } A^c \text{ is finite}\\ 1 & \text{ otherwise }\end{cases}$$
P is finite aditive but is not countably aditive