I know you wanted a reference, but let me try to sketch a proof anyway.

In general, suppose we have an adjoint pair of functors between triangulated categories (or better, cocomplete pretriangulated dg categories or similar(?)).

$$ R: \mathcal C \leftrightarrows \mathcal D: L$$

where $RL \simeq id_D$ (so $L$ is fully faithful).

**Claim (well known, I guess)**: we have a semiorthogonal decomposition $ (ker(R), im(L))$.

**Proof**: First, it is clear that $ker(R)$ is contained in the right orthogonal to $im(L)$ by adjunction. On the other hand, if $M$ is contained in the right orthogonal to $ker(R)$, then consider the map $f:LR(M) \to M$ given by the counit of the adjunction. The map $R(f)$ is an isomorphism as $RL\simeq id$, thus the cone of $f$ is in the kernel of $R$. As $M$ is right orthogonal to $ker(R)$, the cone is zero, so $f$ must be an isomorphism. Thus $M$ is contained in the image of $L$. $\square$

Using the notation from the question, we take $\mathcal C=D(X)^G$, $\mathcal D = D(Y)$, $L=p^\ast$ and $R$ is the functor of invariants. Your question is then reduced to:

**Claim**: $i_\ast D(Z) = ker(R)$.

**Proof**: Everything is local, so we may assume $X=Spec(A)$ is affine, then $Y=Spec(A^G)$, and $Z=Spec(A/I)$, where $I$ is the ideal generated by $g(f)-f$, where $f\in A$, and $g$ is the involution. The functor $L= A\otimes_{A^G} (-)$, and $R = (-)^G$.

It is clear that $i_\ast D(Z)$ is contained in the kernel of invariants. To see the other containment, we must prove that if $M$ is in the kernel of invariants then it is annihilated by the ideal $I$. Note that $g(m)=-m$ for any $m\in M$. Consider

$$ (g(f) - f)(g(m)-m) = 2(g(f)-f)m$$

for $f\in A$ and nonzero $m\in M$. This is a $G$-invariant element of $M$, so must be zero, thus $m$ is annihilated by $g(f)-f$ as required.