ThereFirst of all, there should be a constraint on the exponent $\tau$ to guarantee certain integrability conditions. Define In this case the constraint reads
$$n_*:=\frac{2(n+\lambda)}{n-2}, $$$$n^*:=\frac{2(n+\lambda)}{n-2}, $$
where $n$ denotes the dimension of the ball. Then $E$ embeds continuously in the Banach space $ X:=L^{n_*}(0,1; r^{\lambda+n-1} dr)$$ X:=L^{n^*}(0,1; r^{\lambda+n-1} dr)$. (For a proof of this I refer to this old paper of mine.)
Define $\tau^*:=n^*-1$. (This is the critical Sobolev exponent.) Assume
$$\tau \leq \tau^*. $$
Set $r:=|x|$, $x\in \mathbb{R}^n$. Use Holder's inequality for the conjugate exponents $n^*$ and $\frac{n^*}{\tau^*}$ and the Sobolev embedding $E\to X$ to verify that
$$\left\vert\int_\Omega r^\lambda |u|^\tau v dx\right\vert \leq C\Vert u\Vert_E\cdot \Vert v\Vert_E. $$
This proves that $Tu$ defines a linear functional on $E$. The dual of $E$ is $H^{-1}$.
The Laplacian defines an isomorphism $\Delta E\to H^{-1}$ with inverse $\Delta^{-1}$. The continuity statement is proved by observing that the map
$$ E\ni u\mapsto |u|^\tau \in L^{\frac{n^*}{\tau^*}}(0,1, r^{\lambda+n-1} dr). $$
is continuous.