Is the criterion SAA of congruent triangles valid? I think  the fundamental criterions for triangles congruence are:

*

*SAS (Side-Angle-Side)

*ASA (Angle-Side-Angle)

*SSS (Side-Side-Side)

But some proofs like this one: https://www.youtube.com/watch?v=DArQTsH6Y1s
Use the SAA (Side-Angle-Angle) which I'm not sure if it is valid. I mean, if we assume that the sum of the measure of all angles are the same in all triangles this is indeed the ASA criterion. But otherwise is invalid.
Is that correct?
 A: SAA is perfectly valid, being equivalent to ASA.  Given two angles, the third is immediately determined, since the sum is 180 deg.
A: SAA is valid even in "neutral/absolute geometry", which takes no stand on the Parallel Postulate—or, equivalently, the Angle-Sum Theorem. (Importantly, Side-Angle-Side itself is a postulate in neutral geometry, because you have to start somewhere when comparing triangles.)
See, for instance, this 1977 The Mathematics Teacher article "Neutral and Non-Euclidean Geometry—A High School Course" (JSTOR link) by Krauss and Okolica. From the article:

[M]ost students believed that the side-angle-angle congruence  theorem is not valid in neutral geometry because its traditional proof uses the  theorem that  the sum of the measures of the angles of a triangle is $180$, which they had been shown is equivalent to the parallel postulate. The following proof shows that, in fact, the SAA theorem is valid in neutral geometry. It depends on the exterior angle inequality theorem of neutral geometry, which states: The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. A proof of this theorem can be found in Prenowitz and Jordan (1965, p. 22).
Given: $\angle A\cong\angle A'$, $\angle C\cong\angle C'$, $\overline{AB}\cong\overline{A'B'}$ 
Prove: $\triangle ABC \cong \triangle A'B'C'$
If $\overline{AC}\cong\overline{A'C'}$, the triangles are congruent by SAS. So assume $\overline{AC}\not\cong\overline{A'C'}$, say, $AC<A'C'$. Choose point $D$ on $\overline{A'C'}$ so that $\overline{A'D'}\cong\overline{AC}$. Draw $\overline{B'D}$. $\triangle ABC\cong\triangle A'B'D$ by SAS. Therefore, $\angle C\cong\angle B'DA'$; so $\angle B'DA'\cong\angle C'$. Now we have a contradiction, since $\angle B'DA'$ is an exterior angle of $\triangle B'C'D$, which is greater in  measure than either nonadjacent interior angle, such as $C'$. Hence, or our original assumption that $\overline{AC}\not\cong\overline{A'C'}$ is incorrect. $\overline{AC}\cong\overline{A'C}$ and $\triangle ABC\cong\triangle A'B'C'$ by SAS. A similar proof follows if $AC>A'C'$.

BTW: Limited JSTOR access is free. (For the duration of the coronavirus pandemic, the limits have been made pretty generous: 100 articles/month!) But, conveniently, SAA is mentioned on the first page of the article, and much of its proof is visible in the article's preview image.
