Let $G, G'$ be smooth algebraic groups over $k$ (absolute Galois group $\Gamma$) which are etale inner forms of each other, that is, there exists an isomorphism $G_{k_s} \cong G'_{k_s}$ and the associated torsor for the automorphism group is in the image of the natural map $H^1(k, G/Z_G) \to H^1(k, \operatorname{Aut}_G)$. My question is about the truth of the following proposition:
Proposition: Each inner isomorphism $G_{k_s} \cong G'_{k_s}$ induces a bijection $H^1(k, G) \xrightarrow{\sim} H^1(k, G')$.
Is this true as stated, or only for pure inner forms? Does it follow if we make any further assumptions on $G$ (affine, reductive, semisimple . . .)?
This is asserted a few places, e.g. the answer here. However, the linked proof here (as in Serre's Cohomologie Galoisienne), only works for pure inner forms, i.e. those corresponding to cocycles that factor through $H^1(k, G) \to H^1(k, G/Z_G) \to H^1(k, \operatorname{Aut}_G)$.
Proof for pure inner forms
The argument, as I understand it, goes as follows.
Proof for pure inner forms: Let $s \in \Gamma \mapsto g_s \in G(k_s)$ be a Galois cocycle in $H^1(k, G)$, with associated cocycle valued in $\operatorname{Aut}_G$ given by conjugation, $s \mapsto \left(h \mapsto g_s h g_s^{-1}\right)$. Let $G'$ be the pure inner form of $G$ defined by this latter cocycle. In particular, $G'(k_s) = G(k_s)$, and the two galois actions are related by $s \cdot'(g) = g_s(s \cdot g) g_s^{-1}$.
We define a map $H^1(k, G) \to H^1(k, G')$ on cocycles now as follows: given $s \mapsto a_s \in G(k_s) = G'(k_s)$ a cocycle for the $G$-Galois action, we can define a cocycle for the $G'$-Galois action by $s \mapsto b_s := a_s g_s^{-1}$. This is indeed a cocycle: $$ b_{s} s \cdot'(b_t) = a_s g_s^{-1} g_s(s \cdot(a_t g_t^{-1})) g_s^{-1} = a_s s \cdot(a_t) \; g_{st}^{-1} = a_{st} g_{st}^{-1}, $$ easily seen to descend to $H^1(k, G)$, and has inverse given by right multiplication by $g_s$ instead. QED.
My Thoughts
In the proof above, if $s \mapsto g_s \in (G/Z_G)(k_s) = G(k_s)/Z_G(k_s)$ is valued in the inner automorphism group, then the expression $s \mapsto b_s = a_s g_s^{-1}$ is only well-defined up to an element of the center, and the failure of the identity $g_s s(g_t) = g_{st}$ at the level of $G$ (which was used in checking $b_s$ was a cocycle) is measured by a 2-cocycle valued in $Z_G$. If this 2-cocycle is not a coboundary, we should not expect to be able to choose a cocycle lifting the $g_s$ to $G(k_s)$.
In other words, we should run into the difficulty that the map $H^1(k, G) \to H^1(k, G/Z_G)$ is rarely surjective, fitting into the LES $$ H^1(k, Z_G) \longrightarrow H^1(k, G) \longrightarrow H^1(k, G/Z_G) \longrightarrow H^2(k, Z_G). $$ For $\operatorname{GL}_n$, for example, $Z_G = \mathbb{G}_m$, and so the Brauer group provides an obstruction to lifting cocycles from $\operatorname{PGL}_n$ to $\operatorname{GL}_n$. I haven't been able to construct a counterexample among the forms of $\operatorname{GL}_n$, however.