Two congruent segments does have the same length? The answer to the question in the title seems an obvious ''Yes by definition !''. And this really is the definition from Wikipedia:

Two line segments are congruent if they have the same length.

But in Hilbert's Foundations of Geometry congruence is defined without use of metric notions, by the axioms of Group IV (chapter 6).
I don't report here these axioms that can be found on the book (or on the Wiki page)
and essentially say that congruence is an equivalence relation and that congruence is conserved when we add adjacent segments.
Usually these axioms are illustrated with a figure that use two segments $AB$ and $A'B'$ that have the same length, as in this case. 

but I think that this illustration is only dictated by our intuition of congruence as equality of length, and it's not justified by the axioms.
Look at the second figure, where $a$,$a'$ and $a''$ are straight lines and $E$ is a point that does not belong to these lines. 

Projecting from $E$ the points $A,B,C$ of $a$ to the other lines, we can define the relation:
$$
AB \equiv A'B' \quad BC \equiv B'C'
$$ 
and
$$
A'B' \equiv A''B'' \quad B'C' \equiv B''C''
$$ 
and we can see that this relation satisfies all axioms of congruence. But, clearly, it doesn't satisfies our intuition that congruent segments have the same length.
This construction is essentially the same that Hilbert uses (in the chapter 24) for the  construction of an ''Algebra of Segments'', ( where the point $E$ is ''at infinity'').
Obviously ''to have the same length'' is a congruence relation, but only in a metric space, and, in the same space, different congruence relations can be defined that are not equivalent to ''have the same length''. And more, we can have different metric (and different ''length'') for a geometry that satisfies the Hilbert's axioms.
So my question is, at first, if my reasoning is correct or if I've misunderstood something of fundamental, and, if I'm right, what are the axioms that we need for exactly represent our intuitive concept of congruence?
A final notice: this question is related to What really is ''orthogonality''?, where a similar question is posed for angles.
 A: First of all, in order to make sense of the question one has to specify the axiomatic system via which the (Euclidean) plane is defined. For the purpose of this answer, I assume that you are interested in Hilbertian axiomatization of the Euclidean geometry. (There are alternatives, such as Birkhoff's and Tarski's sets of axioms.) Hilbert's axiomatization has two equivalence relations built in: congruence of segments and congruence of angles. If you want to change Hilbert's notion of congruence to something else, you are leaving the realm of Hilbert's axioms and, hence, you would have to list your personal axioms and then ask a new question. 
Given this, the answer is that, by the definition, two segments have the same length iff they are congruent, the congruence class of a segment $AB$ is denoted $[AB]$. 
This, of course, is not entirely satisfactory since we are used to the idea that length should be a real number and not an element of some unfamiliar quotient set (coming from some equivalence relation). I suspect that this is where your question is coming from. 
It turns out that one can (assuming all but the continuity axioms of Hilbert) define an ordered field $F$ whose elements are the congruence classes, you can find a detailed construction in the book: 
R. Hartshorne, "Geometry, Euclid and Beyond", Springer Verlag, 2000. 
One needs the two continuity axioms (of Hilbert's axiomatic system) in order to prove that $F$ is isomorphic to the field of real numbers.  
Note. In contrast, if you accept Birkhoff's axiomatization, then the Euclidean plane $E$ comes equipped with a distance function $d: E\times E\to {\mathbb R}_+$. The length of a segment $AB$ is simply the real number $d(A,B)$. A congruence of this geometry is an isometry, i.e. a map $g: E \to E$ preserving the distance function. Two segments $AB, CD$ are congruent in this sense iff there exists an isometry $g$ such that $g(AB)=CD$. Thus, two congruent segments have the same length, again, just by the definition. A good reference for this is: 
E. Moise, "Elementary Geometry from an Advanced Standpoint", Pearson, 1990. 
Moise sketches the relation between the two (Hilbert and Birkhoff) sets of axioms and explains, for instance, how to add congruence classes in Hilbert's plane. But he does not verify any of the ordered field axioms (for instance, he does not explain how to multiply the congruence classes). 
Edit. I think, I understand now the source of your confusion, it is hidden in the article an. The point is that Hilbert's axiomatic system does not just contain an equivalence relation, called congruence, but a very specific one, which you can find for instance in either one of the references I gave above. If you change Hilbert's segment congruence equivalence relation to a different equivalence relation, you obtain a different axiomatic system. It appears that you are interested in replacing the standard segment congruence equivalence relation  with the similarity equivalence relation. This equivalence relation has exactly two equivalence classes, namely, one represented by degenerate segments $AA$ and one represented by nondegenerate segments $AB$, $A\ne B$. This is all fine, but now you are asking about "length" in relation to this equivalence relation. For this, you have to specify what do you mean by the "length" since the latter is not a part of Hilbert's axiomatic. There are two meaningful choices:
a. Use the notion of length $[AB]$ as the equivalence class of the segment $AB$ with respect to the standard notion of congruence. This notion is clearly not preserved by the similarity: Similar segments, in general, do not have the same length. This is quite immediate (you can use the fact that there are infinitely many congruence classes of segments in the traditional sense and only two similarity equivalence classes). Thus, the answer to your question is negative in this case. 
b. Use the notion of "length" of $AB$ as the similarity equivalence class of the segment $AB$. Then your notion of "length" takes exactly two values and is, but the definition, preserved by the similarity equivalence relation. Very few (if any) people, however, will accept this notion of "length" as useful, for instance, because it does not take values in the ordered field (the field of two elements cannot be ordered). 
A: I think everyone is ignoring a crucial fact.  Hilbert's  congruence axiom for segments is a axiom about an undefined word: congruence. (By congruence, I mean the congruency of segements although the argument I'm presenting may apply to the congruency of angles too.) Hilbert purposely left the word congruence undefined so that an interpretation may be given by anyone who pleases. Surely, you know that an interpretation of congruence must be associated with some model  that someone is trying to show satisfies Hilbert's congruence axioms based upon its interpretation of congruence. 
This is how models for axiomatic systems work to begin with.  In order for there to be a model for a set of axioms, the axioms must contain undefined terms for which the model provides interpretations and the axioms must also be true under this  interpretation. 
The interpretation of congruent segments by means of the idea that two segments are the same length may be provided by any metric space that has a defined metric which I believed you mentioned. But this merely provides a model for Hilbert's axiom, in other words, an interpretation. It does not replace the axiom!
The example in which you gave another interpretation of congruence without the intuitive idea of length is just that: just another interpretation or model. It again does not replace the axiom, but provides an interpretation for the word congruence.
