The claimed triviality holds (the nonabelian cohomology set is not a group though), and I don't think you need $Char(K) \neq 2$. To argue this, I will use the long exact nonabelian cohomology sequence of the central extension $1 \rightarrow \mathbf{G}_m \rightarrow GL_2 \rightarrow PGL_2 \rightarrow 1$, a segment of which reads $H^1(K, GL_2) \rightarrow H^1(K, PGL_2) \rightarrow H^2(K, \mathbf{G}_m)$. Firstly, $H^1(K, GL_2)$ is the one-point set because it classifies rank 2 vector bundles over $Spec(K)$, of which there is only the trivial one. Secondly, $K$ is a $C_1$ field by a theorem of Lang (see Serre "Galois cohomology", p. 80, II.3.3 c)), hence is of $dim \le 1$, so its Brauer group $H^2(K, \mathbf{G}_m)$ vanishes (loc. cit. for more details).

The same argument shows that $H^1(K, PGL_n)$ is the one-point set for any $n$ and any $C_1$ field $K$.