Is the Euclidean plane equal to $\mathbb{R}^2$? I understand what the Cartesian plane is and how one can identify the Euclidean plane and $\mathbb{R}^2$ in many different ways. My question comes down to this though:

The Euclidean plane (as defined in the link) has no distinguished points, whereas $\mathbb{R}^2$ very clearly has a distinguished point in the origin $(0,0)$.
Therefore, doesn't it only make sense to identify $\mathbb{R}^2$ with the tangent spaces of the Euclidean plane, which very clearly do have distinguished points?

One often says that $\mathbb{R}^n$ is "Euclidean space", which admittedly does make sense from the viewpoint of analysis, algebra, and topology, among others. However, it seems like from a geometric and a categorical viewpoint, saying that the two are the same could only lead to confusion.
A similar, but certainly not identical view, seems to be voiced by Professor Lee in his book Introduction to Smooth Manifolds, p. 51 of chapter 3:

...We have to come terms with a dichotomy in the way we think about elements of $\mathbb{R}^n$. On one hand, we usually think of them as points in space, whose only property is location, expressed by the coordinates $(x^1, \dots, x^n)$. On the other hand, when doing calculus we sometimes think of them instead as vectors, which are objects that have magnitude and direction, but whose location is irrelevant. A vector $v=v^ie_i$ (where $e_i$ denotes the $i$th standard basis vector) can be visualized as an arrow with its initial point anywhere in $\mathbb{R}^n$; what is relevant from the vector point of view is only which direction it points and how long it is. What we really have in mind here is a separate copy of $\mathbb{R}^n$ at each point. When we talk about vectors tangent to the sphere at a point $a$, for example, we imagine them as living in a copy of $\mathbb{R}^n$ with its origin translated to $a$.

EDIT: To clarify, $\mathbb{R}^2$ should in theory mean just a pair of reals, however this is not what my question refers to. In practice, people seem to refer to "$\mathbb{R}^2$ with the standard inner product, the standard norm which is the norm induced by the standard inner product, the standard metric which is the metric induced by the standard norm, and the standard topology which is the topology induced by the standard metric" as "$\mathbb{R}^2$", and to call of these structures X the "Euclidean X", i.e. the "Euclidean inner product", "Euclidean norm", "Euclidean metric", "Euclidean topology", and then call $\mathbb{R}^2$ itself the "Euclidean plane". Ideally there would be different terms for both of these objects, but in practice there does not seem to be. My question is about the difference between the Euclidean plane (of Euclid) and "$\mathbb{R}^2$" as conceived of in elementary textbooks in linear algebra and real analysis, where all of the standard additional structures which could be defined on the space are assumed to be defined on the space. END EDIT
For instance, the Euclidean plane isn't a vector space without specifying an origin, in which case one is essentially identifying it with its tangent space at that point. Moreover, keeping this distinction in mind makes it clear why linear and affine transformations should be considered different, since the latter requires moving between different tangent spaces of the plane. Also, Euclidean space clearly isn't an inner product space either, because given two points of Euclidean space, their angle is not defined (in contrast to the angle between two non-zero elements of $\mathbb{R}^2$). The angle between two lines or two line segments is defined only when they intersect at a distinguished point with respect to which the angle is calculated. Not incidentally the angle between vectors in the tangent space based at that point corresponding to the given lines or line segments equals the same angle as defined in Euclidean space -- i.e. one can use the metric space structure of Euclidean space to define a Riemannian metric on its tangent spaces.
I know that when $\mathbb{R}^2$ is considered as an algebraic group (e.g. addition is continuous, translation is a homeomorphism) results for neighborhoods at $(0,0)$ magically apply for neighborhoods at every point, and thus we can show that $\mathbb{R}^2$ with such a topology is homeomorphic to the Euclidean plane. However that doesn't mean that addition of points is a priori well-defined in the Euclidean plane, since there is no natural choice for the additive identity element, unlike in $\mathbb{R}^2$. Only the tangent spaces of the Euclidean plane have natural choices for an additive identity element.
Note: Almost the exact same question has been asked, but in greater generality, here. However, the only answer to that question "it's a matter of choosing a coordinate system" was not accepted and I also find it unsatisfactory -- if $\mathbb{R}^2$ comes with a standard coordinate system, but the Euclidean plane does not, then clearly they are not the same object, at least they clearly are not isomorphic in at least some category. The answer to this question comes closer to being precise, but it still seems to waffle somewhat, and hence was also unaccepted. This similar question was unanswered.
If one is concerned about the construction of the Euclidean plane without using $\mathbb{R}^2$, there exists an axiomatization of the object in the framework of absolute geometry plus the appropriate parallel postulate for Euclidean space, which does not start from $\mathbb{R}^2$, for example, see Elementary Geometry by Agricola and Friedrich. From a geometric point of view it seems to make more sense to think of the plane this way, because then its geometry is "intrinsic", rather than being the result of being embedded in $\mathbb{R}^2$, and thus better motivates the idea of abstract manifolds existing in their own right without necessarily having to be embedded in $\mathbb{R}^n$, as well as the ideas that the geometry of the hyperbolic or projective plane should be on an equal footing with that of the Euclidean plane.
 A: "The" Euclidean plane does not exist. Rather, anything that fulfils the Euclidean plane axioms is an Euclidean plane. Now whatever fulfils those axioms may have additional structure (for example, the distinguished point $(0,0)$ of $\mathbb R^2$). But that additional structure is not part of its Euclidean plane structure.
It's the same with most constructions. For example, "the" natural numbers may be constructed using sets. With that construction, we can e.g. calculate the set difference of the set representing $5$ and the set representing $3$, despite the fact that $5\setminus 3$ makes no sense for natural numbers (and indeed, at least in the canonical construction, the result is not a set representing a natural number). But that is not a deficiency of the construction; indeed, if we want to construct numbers from sets, we cannot avoid this.
The point is, when treating your construction as model of some axiom set, you ignore any additional structure it may have. When treating $\mathbb R^2$ as Euclidean plane, you ignore that you've got a preferred point $(0,0)$ and two preferred directions $(x,0)$ and $(0,y)$. And when treating certain sets as natural numbers, you ignore that those sets have set properties, and you simply don't ask for things like the symmetric difference.
More generally, if you simply never ask for things that cannot be asked for in the language of the axiom system, the extra structure will be "invisible" to you, and therefore can be safely ignored.
It's actually not that different to the fact that when speaking about the properties of numbers, you ignore e.g. that the digit 6 has a closed loop, while the digit 2 has none. It's true, but irrelevant to what you are talking about.
A: I'd say that no, the Euclidean plane is not equal to $\mathbb R^2$. But $\mathbb R^2$ with standard inner product and so on and so forth is a model of the Euclidean plane. One can translate concepts in the Euclidean plane into corresponding concepts in $\mathbb R^2$, so e.g. a point becomes a vector.
It is a model which comes with additional structure (like e.g. a designated point of origin, or the class of linear transformations) which has no counterpart in the axiomatization of the Euclidean plane, but that doesn't prevent it from being a model.
Since models are so very useful, many users may not bother to always draw a clear line between the abstract theory and their model of choice. So for many applications, thinking of the Euclidean plane and $\mathbb R^2$ as essentially the same may well represent how people think about it.
A: I think mathematicians sometimes use the word 'is' in a slightly confusing manner, namely to mean 'isomorphic'; once you get used to it, it is actually a help to 'free up' your mind, in the sense that it lets you disregard the inessential differences: when two objects of a given kind (actually: in a given category) are isomorphic, they are 'the same' for the purposes of whatever you are discussing.
So, in the case of the/a Euclidean Plane, it is isomorphic to $\mathbb{R}^2$, if you equip both both a vector space structure - and of course, when you do so, you automatically select a point of origin in the plane. This does not lead to any loss of generality in most cases, since you can choose any point as your origo; and while the Euclidean Plane regarded in isolation does not define a distinguished point, in practice you impose one as soon as you start on a geometric construction.
