= Solution
First use the usual <Hilbert space> convention for the <proximal operator>, identifying the dual element $z$ with its representing vector. For $c\ne0$, completing the square gives
$$
\frac12\|x-q\|^2+\langle x,z\rangle=\frac12\|x-(q-z)\|^2+\text{constant}.
$$
Set $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\|v-[c(q-z)-y]\|^2/2+\alpha c^2J(v)$. Thus the <proximal operator under affine rescaling> is
$$
\boxed{\operatorname{prox}_E(q)=\frac{y+\operatorname{prox}_{\alpha c^2J}(c(q-z)-y)}c\qquad(c\ne0).}
$$
Negative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.
The PDF actually assumes an arbitrary <Banach space>, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define $\operatorname{Prox}_F(s)=\arg\min_v\{F(v)+\|v-s\|^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the <norm> still gives the valid change of variables
$$
\boxed{\operatorname{Prox}_E(q)=\frac{y+\operatorname{Prox}_{\alpha c^2J+c\langle\cdot,z\rangle}(cq-y)}c\qquad(c\ne0).}
$$
This retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the <duality mapping> $\mathcal J=\partial(\|\cdot\|^2/2)$:
$$
0\in\mathcal J(x-q)+\alpha c\,\partial J(cx-y)+z.
$$
For a concrete failure of the Hilbert shortcut, take $(\mathbb R^2,\|\cdot\|_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\leq0$ with $x_1+x_2=-1$ minimizes $\|x\|_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence \b[the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique]. For $c=0$ with finite constant penalty the Banach condition is $-z\in\mathcal J(x-q)$, not generally $x=q-z$.
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