(1) The flat plane always has isometry group $ E_2 = O_2 \ltimes \mathbb{R}^2 $.
(2) A full classification of the four possible isometry groups of a flat torus are given in
Isometry of Torus
The isometry group of a product of (irreducible, non isometric) Riemannian manifolds is the product of the isometry groups. This point is basically addressed in How does one prove that $\text{Isom}(\mathbb{S}^2 \times \mathbb{R}) = \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$? and https://mathoverflow.net/questions/351646/the-isometry-group-of-a-product-of-two-riemannian-manifolds so
(3) So the isometry group of a flat cylinder $ \mathbb{R} \times S^1 $ is always $ Iso(\mathbb{R}) \times Iso(S^1)= E_1 \times O_2=(O_1 \ltimes \mathbb{R} ) \times O_2 $
Now my question: is a full classification of isometry groups, as done for the torus, possible for the other flat surfaces:
(4) The flat Moebius band (I think the isometry group is always $ Iso(\mathbb{R})=E_1=O_1 \ltimes \mathbb{R} $, just isometries of the fiber)
(5) The flat Klein bottle (I think the isometry group is always $ Iso(S^1)=O_2 $, just isometries of the fiber)
EDIT: Using the hint from Kajelad and applying it to coverings of both of these manifolds by the flat cylinder I compute the following.
The Moebius strip is the quotient of the cylinder by an isometry acting on points in the cylinder $ (t,z) \in \mathbb{R}\times U_1 $ by
$$
(t,z) \to (t,\overline{z})
$$
this generates a two element cyclic group $ \Gamma $ of isometries which intersects two of the connected components of the isometry group $ (O_1 \ltimes \mathbb{R} ) \times O_2 $ of the flat cylinder. Since $ \Gamma $ has only two elements the normalizer is the same as the centralizer. And the centralizer is the full isometry group of the line together with the isometries of the circle $ z \to -z $, $ z \to \overline{z} $ and product. So modding out $ N(\Gamma)/\Gamma $ is the isometries of the fiber (line) direct product a cyclic two group acting on the base (circle) so the full isometry group of the flat Moebius strip is $ O_1 \ltimes \mathbb{R} ) \times O_1 $.
The Klein bottle is the quotient of the cylinder by an isometry acting on points in the cylinder $ (t,z) \in \mathbb{R}\times U_1 $ by
$$
(t,z) \to (t+1,\overline{z})
$$
this generates an infinite cyclic group $ \Gamma $ of isometries which intersects two of the connected components of the isometry group $ (O_1 \ltimes \mathbb{R} ) \times O_2 $ of the flat cylinder. The normalizer of $ \Gamma $ is a group of isometries consisting of all the translations in the $ \mathbb{R} $ factor together with the isometries $ z \mapsto \overline{z} $ and $ z \mapsto -z $ in the circle factor. The quotient $ N(\Gamma)/ \Gamma $ is isomorphic to $ O_2 $ (since $ \Gamma $ intersects two different connected components it cuts down the four connected components of the normalizer to just two in the quotient and since $ \Gamma $ is cyclic it cuts down the $ \mathbb{R} $ translation factor in the normalizer to just a compact piece $ \mathbb{R}/\mathbb{Z} $ in the quotient. So in conclusion the isometry group of the Klein bottle is just $ O_2 $ the isometries of the circle fiber.
Best Answer
If you're familiar with covering spaces, then there's an approach which should be useful for each of these:
Given a Riemannian manifold $(M,g)$, the universal cover $\widetilde{M}$ can be equipped with a metric $\widetilde{g}$ such that the covering map $\widetilde{M}\mapsto M$ coincides with the quotient map $\widetilde{M}\mapsto\widetilde{M}/\Gamma$ where $\Gamma$ is a discrete subgroup of isometries of $\widetilde{M}$. Using among other things the lifting property of universal coverings, you can show that the isometry groups are related by the following expression: $$ \operatorname{Isom}(M)\cong N_{\operatorname{Isom}(\widetilde{M})}(\Gamma)/\Gamma $$ (here $N$ denotes the normalizer)
Each of the listed manifolds are universally covered by $\mathbb{R}^2$, so they can all be described as quotients of the Euclidean plane by discrete subgroups of $E(2)$. Actually computing these will depend on which flat metrics you're interested in.