Do finitely additive measures solve the "problem of measure"?

Question

Does there exist a unique function $\mu$ satisfying the following properties?

1. $\mu:\mathcal P(\mathbb R)\to [0,\infty]$
2. $\mu(A+x)=\mu(A) \qquad\qquad$ for all $A\in\mathcal P(\mathbb R),x\in\mathbb R$
3. $\mu([a,b])=b-a\qquad\qquad\,$ for all $a<b\in\mathbb R$
4. $\mu\big(⨃_{i=1}^N A_i\big)=\sum_{i=1}^N \mu(A_i)\;\,$ for all finite pairwise-disjoint families $(A_i)$ with each $A_i\in\mathcal P(\mathbb R)$

If (4) is strengthened to include countably infinite disjoint unions, such a function famously does not exist (which Terrence Tao has called the "problem of measure," hence the question title). The standard solution is to replace every occurence of $\mathcal P(\mathbb R)$ above with $\mathcal B(\mathbb R)$; then Borel measure is the unique function on $\mathcal B(\mathbb R)$ that satisfies the above conditions (with 4 strengthened to countable additivity).

I'm curious whether this alternative approach (using only finitely additive measures) would work instead.