Why is Euclidean geometry scale-invariant?

Maybe this reason isn't "deep" enough, but it's because the Euclidean distance formula (the Pythagorean identity) is scale invariant:

$$a^2 + b^2 = c^2$$

$$(Xa)^2 + (Xb)^2 = X^2(a^2+b^2) = (Xc)^2$$

whereas the distance formulas in spherical geometry:

$$\cos a \cos b = \cos c$$

and hyperbolic geometry:

$$\cosh a \cosh b = \cosh c$$

lack this property.

Similarity is a special feature of Euclidean geometry. It follows from the Euclidean parallel postulate:

For each line and each point not on the line, there is a unique second line, parallel to the first, that contains the point.

As a matter of fact, the following claim, called Wallis's postulate, is equivalent to the parallel postulate:

Given any triangle $\triangle ABC$ and any scale factor, r, there exists a triangle $\triangle DEF$ similar to $\triangle ABC$ where r is the ratio of the lengths of corresponding sides.

There is a list of 15 claims all logically equivalent to the Euclidean parallel postulate on the Wikipedia Parallel Postulate article. One of which, not surprisingly, is the Pythagorean theorem.

Perhaps the following will provide some insight.

Assume

AAA Similarity Theorem: If two triangles have all three pairs of angles congruent, then the triangles are similar.

the proof of which does not require the parallel postulate but is rather involved. Nonetheless, with it, we can show:

Similar Triangle Construction Theorem: If $\triangle ABC$ is triangle and $\overline{DE}$ any segment, then there is a point $F$ such that $\triangle ABC$ is similar to $\triangle DEF$.

Proof: We can construct a unique ray $\overrightarrow{DP}$ so that $\angle EDP \cong \angle A$. We can similarly construct a unique ray $\overrightarrow{EQ}$ on the same side of $\overleftrightarrow{DE}$ so that $\angle DEQ \cong \angle B$. Since the angle measures of the constructed angles, $\angle EDP$ and $\angle DEQ$, sum to less than $180$, by Euclid's fifth postulate (equivalent to Euclidean parallel postulate), the two rays must intersect at some point $F$ forming a triangle $\triangle DEF$. By the angle-sum theorem (also logically equivalent to the Euclidean parallel postulate), $\angle C \cong \angle F$. Now by the AAA similarity theorem, the two triangles are similar. $\square$

Then given any scale factor $r$, We can construct a segment $\overline {DE}$ so that $$r=\frac{DE}{AB}.$$

From this, Wallis's postulate follows straight forwardly.

Conversely, let us assume Wallis's postulate and from it derive the Euclidean parallel postulate.

Proof of EPP from Wallis's Postulate: Suppose we have a line $\ell$ and a point $P$ not on the line. We can drop a perpendicular from $P$ to line $\ell$ and let the perpendicular intersect $\ell$ at point $F$. Then construct a new line $m$ at $P$ at $90$ degrees to our perpendicular. This new line will be parallel $\ell$. If it were not, $m$ and $\ell$ would intersect forming a triangle which would violate the exterior angle theorem.

It remains to show the line $m$ is unique. Suppose there were a different line $n$ also through $P$. We must show $n$ is not parallel, that is, line $n$ intersects $\ell$.

If $n$ is our perpendicular $\overleftrightarrow{PF}$ obviously it intersects $\ell$. So if $n$ is not $\overleftrightarrow{PF}$, lets choose a point $Q$ of $n$ between lines $\ell$ and $m$ but not on $\overleftrightarrow{PF}$. Let $R$ be the point of intersection of a perpendicular dropped from $Q$ onto line $\overleftrightarrow{PF}$. Let $r=\frac{PF}{PR}$. Wallis's theorem assures us that there is somewhere a triangle $\triangle ABC$ that is similar to $\triangle PQR$ with scale factor $r$. We can construct a new triangle $\triangle PSF$ on the segment $\overline {PF}$ congruent to $\triangle ABC$ by placing $S$ on $\ell$ at distance $BC$ from $F$ and on the same side of $\overleftrightarrow{PF}$ as $Q$. This triangle is congruent to $\triangle ABC$ by the side-angle-side congruence theorem and so is similar to $\triangle PQR$. So the angles $\angle FPS$ and $\angle RPQ$ are the same angle. Hence the lines $n$, $\overleftrightarrow{PQ}$, and $\overleftrightarrow{PS}$ are the same line, and $n$ does indeed intersect $\ell$. $\square$