The question is not well-posed. There are various versions of cyclic theory (for instance) which differ according to continuity conditions that are assumed. In Connes' original IHES papers he deals with both discrete (useful for arbitrary rings) and topological (useful in the $C^\infty $-setting.)
The basic problem is that cyclic theory is very very sensitive. Consider the following example (using Connes' topological cyclic theory.) Let $M$ denote a compact smooth manifold with a smooth foliation. Then there are three operator algebras that you can associate with the situatio:;
a) $C(M)$, the $C^\ast$-algebra of continuous complex functions on $M$.
b) $C^\infty (M)$, the smooth functions on $M$.
c) $C_\tau ^\infty (M)$, the continuous functions on $M$ which are smooth in the leaf directions.
The cyclic cohomology of these three rings are all different typically (e.g. for the Kronecker flow on the torus).
a) leads to measures on $M$;
b) leads to de Rham cohomology on $M$;
c) leads to "tangential cohomology" on $M$ (cf. my book Global analysis on foliated spaces with Cal Moore).
Best Answer
Here is a down-to-Earth counter-example. Set $N=\mathbb C$ and $M=\mathcal B(K)$ with $H=K\otimes K$ and $M$ acting on the first tensor factor (should perhaps be $K\otimes\overline K$ but this will not affect the argument). Let $K$ have orthonormal basis $(e_n)$ and take for example $\xi = \sum_n n^{-1} e_n\otimes e_n$.
Then $\xi$ is also separating, and so the module map sending $\xi$ to $\eta$ will exist for any $\eta$. (In general, this need not be the case I think, so already in complete generality the bimodule map sending $\xi$ to $\eta$ might not be well-defined.) I now make some choices: let $$ \eta = \Big( \sum_n n^{-3/4} e_n \Big) \otimes e_1. $$ Consider the matrix unit $e_{m,n}$ which sends $e_n$ to $e_m$, so our operator is $$ T:e_{m,n}\cdot \xi = n^{-1} e_m \otimes e_n \mapsto e_{m,n}\cdot\eta = n^{-3/4} e_m \otimes e_1. $$ Hence $T:e_m \otimes e_n \mapsto n^{1/4} e_m \otimes e_1$. Consider the sequence $(e_1\otimes\alpha_k) = (k^{-1/4} e_1\otimes e_k) \rightarrow 0$ in $H$, while $$ T(\alpha_k) = k^{-1/4} k^{1/4} e_1\otimes e_1 = e_1\otimes e_1, $$ for all $k$. Thus $T$ is not closable.