The following theorem is well known:
There exists a isometry invariant finitely additve measure, measuring all subsets of $\mathbb{R}^d$ that extends the Lebesgue measure if and only if $d\le 2$
The fact that no such measure exists in $\mathbb{R}^d$ with $d>2$ is the famous Banach-Tarski paradox. The fact that there exists a such a measure with $d=1,2$ was first proven by Banach.
Using the semi-group invariant version of the Hahn-Banach theorem, (see Exercise 2.15 here), one can prove that there exists such a measure in $\mathbb{R}$. I am wondering how to quickly extend this to $\mathbb{R}^2$, doing a little research, I found Wagon's paper "Invariance properties of finitely additive measures in $\mathbb{R}^n$", in that paper Wagon claims that Banach extended the result to $\mathbb{R}^2$ using a "clever integration trick", can someone please spell out the details of this "trick"? I can not read Banach's original paper because it is in French.
[Warning, my mathematical French is very rusty, so any errors below are certainly mine and not Banach's, and my discussion of the "intuition" for the construction is not in the original paper and might be wrong.]
Banach first constructs not just a finitely additive and translation invariant measure on $\mathbb{R}$ that extends Lebesgue measure, but also a finitely additive and translation invariant linear functional $L$ on $B(\mathbb{R})$, the set of bounded functions on $\mathbb{R}$ with bounded support, that extends integration with respect to Lebesgue measure (where as usual you get the set measure by applying the linear functional $L$ to indicator functions of sets).
Having constructed such a thing, one can very loosely think of $L(f)$ as something akin to "$\int_{-\infty}^{\infty} f(t) \, dt$," as long as one does not take this notation too seriously (e.g., if one regards it as perhaps suggestive of the finite additivity and translation invariance properties that $L$ does have, but not as suggestive of other properties that actual integration with respect to a countably additive measure might have). With such a map $L$ on $B(\mathbb{R})$, Banach then constructs a linear functional on $B(\mathbb{R}^2)$ that extends integration with respect to planar Lebesgue measure $A$ on bounded sets, and is invariant under (pre-)composition with isometries of $\mathbb{R}^2$.
The idea is fairly simple. Given $f \in B(\mathbb{R}^2)$, by analogy with the expression of $\iint f \, dA$ (i.e., the thing we are trying to extend, which makes sense if $f$ is Lebesgue integrable) in terms of iterated integrals, two numbers suggest themselves as candidates for an extension of $L$ to $\mathbb{R}^2$, namely $$ I_1(f) := L(x \mapsto L(y \mapsto f(x,y))), $$ the analog of $\int_{-\infty}^{\infty} (\int_{-\infty}^{\infty} f(x,y) \, dy ) \, dx$, and $$ I_2(f) = L(y \mapsto L(x \mapsto f(x,y))), $$ the analog of $\int_{-\infty}^{\infty} (\int_{-\infty}^{\infty} f(x,y) \, dx )\, dy$. And indeed, thanks to the additivity and translation invariance of $L$, both $I_1$ and $I_2$ are finitely additive and translation invariant extensions of integration with respect to Lebesgue measure.
Letting $R$ denote "reflection about the line $y=x$," i.e. the map $(s,t) \mapsto (t,s)$, there is no obvious formal reason why either $I_1$ or $I_2$ would be invariant under $R$. But we do have that $I_1 \circ R = I_2$ and $I_2 \circ R = I_1$, so that the average $I_0(f) = \frac{1}{2} (I_1(f) + I_2(f))$ will be invariant under $R$, and also invariant under translations, because $I_1$ and $I_2$ are. (The use of the average and not the sum or some other combination is to keep the thing normalized so that it has a chance of extending integration with respect to Lebesgue measure, and does not e.g. become a multiple of Lebesgue measure.)
Letting $\rho_{\theta}$ denote "counterclockwise rotation by $\theta$," for each $\theta \in [0, 2\pi]$, i.e. $\rho_{\theta}: \mathbb{R}^2 \to \mathbb{R}^2$ is given by $(s,t) \mapsto (s \cos\theta - t \sin \theta, s \sin \theta + t \cos\theta)$, there is no obvious formal reason why $I_0$ would be invariant under $\rho_{\theta}$. But if we were to let $f_{\theta} := f \circ \rho_{\theta}$ for each $\theta \in [0, 2\pi]$ and define $I$ by $$ I(f) = \frac{1}{2\pi} L(\theta \mapsto I_0(f_{\theta})), $$ then $I$ will be invariant under each of the maps $\rho_{\theta}$, and $I$ also inherits the translation and reflection invariance properties of $I_0$. (Similar to as before, the normalization $\frac{1}{2\pi}$ keeps the thing a fixed multiple of Lebesgue measure.)
Thus $I$, being invariant under translations, a reflection, and rotation about the origin, is invariant under the full Euclidean group.
So that's Banach's construction (my notation is slightly different and he puts the $2\pi$ normalization somewhere else). There are maybe some details to check here, e.g. that $I$ really does extend integration with respect to planar Lebesgue measure (this is clear enough to me for rectangles, or even products of measurable sets, but the fact that $I$ is only finitely additive maybe means there is more to be said to get it for all sets).
I admit that I don't immediately see what goes wrong with this type of idea in $\mathbb{R}^3$, although I have not thought too long about why. It seems easy to create functionals on $B(\mathbb{R}^3)$ that are invariant under translation and at least some reflections; the hard part might be making it invariant under rotations too (this certainly makes sense, knowing how examples of the Banach Tarski paradox are constructed). It certainly seems important in the above that the entire special orthogonal group of $\mathbb{R}^2$ is a group of translations that we can average over with one application of our map from the one-dimensional case. It may be that the obstacles involved in $\mathbb{R}^3$ relate at a high level to the structure of three dimensional rotations simply being 'more complicated than that.'