I don't believe so, here is my proposition.
We know
$$\|T_yf\|_{L^p}=\|f\|_{L^p}$$
This is because the Lebesgue Measure is translationally invariant.
i.e.
$$E\in \mathcal{M}\implies \mu(x+E)=\mu(E)$$
Where $\mathcal{M}$ is the sigma algebra of measurable sets.
This does not hold for all measures.
So in our case let's consider the following counterexample.
Let $E\in \mathcal{M}$ and say $\mu(E)<\infty$ and let's also assume that E is a bounded set.
Consider $\chi_E$ the characteristic function defined on E.
I think we can agree that
$$\chi_E\in L^p$$ and $$\chi_E\in L^p_w$$
If we apply $T_y$ to our function then $\|\cdot\|_{L^p_w}$ will not be preserved because our weight varies with position and so the measure of E will not be translation invariant.
This is just what my intuition says, perhaps someone else can elaborate a little more.
Take care.