Is there exists such figure $A$ on plane $E^2$ which isometry group is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ I suppose that there are no such $A \subset E^2$ which satisfy
$$\text{Iso}(A) \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$
But I'm stuck on showing this in formal way. Can we use here Classification of Euclidean plane isometries theorem to prove it ?

One way to prove this is to prove that the isometry group of $E^2$ does not even contain a subgroup isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$. Yes, the classification of Euclidean isometries will help, but you also need to know about some special subgroups of the group of isometries. In particular, you can use the following facts:


*

*For every finite group of isometries $G$ of $E^2$, there exists a point $x \in E^2$ fixed by each element of $G$.


So we can assume that your subgroup $G = \text{Iso}(A) \approx \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$ fixes some point $x$.


*

*The subgroup $\Gamma_x < E^2$ of all isometries that fixes $x$ can be decomposed as a split short exact sequence
$$1 \to S^1_x \to \Gamma_x \to \mathbb Z_2 \to 1
$$
where $S^1_x$ is the circle group of rotations around $x$, and a splitting $\mathbb Z_2 \to \Gamma_x$ is given by any reflection across a line through $x$.


So, $G$ is contained in $\Gamma_x$.


*

*The normal subgroup $S^1_x$ contains a unique order $2$ element, namely the $180^\circ$ rotation around $x$. That element generates an order 2 subgroup $R_x$, which is a normal subgroup of $\Gamma_x$.


To finish the proof we consider two cases. 
If $G$ contains $R_x$ then $G \cap S^1_x = R_x$, because of uniqueness of the order $2$ element in the group $S^1_x$. It follows that the homomorphic image of $G$ in $\mathbb Z_2$ is isomorphic to $G/R_x$ which has order $4$. 
Whereas if $G$ does not contain $R_x$ then $G \cap S^1_x$ is trivial, by the same uniqueness argument. It follows the homomorphic image of $G$ in $\mathbb Z_2$ is isomorphic to $G$ which has order $8$. 
In either case one gets a contradiction: $\mathbb Z_2$ only has order 2.
