I am trying to show for a separable Hilbert space $H$ with onb $e_n$ $$\Phi: \ell^\infty \to B(H),\quad (a_n)\mapsto \left(x\mapsto \sum_{n\in \mathbb{N}} a_n \langle x,e_n\rangle e_n\right)$$ is a ring homomorphism but getting stuck.
$$\Phi(1_n)(x)=\sum_n\langle x,e_n\rangle e_n=x$$ and $$\Phi(a+b)(x)=\sum_n(a+b)_n\langle x,e_n\rangle e_n=\sum a_n\langle x,e_n\rangle e_n+\sum b_n \langle x,e_n\rangle e_n=\Phi(a)x+\Phi(b)x$$
but for multiplicativitly $$\Phi(ab)(x)=\sum_{n}(ab)_n\langle x,e_n\rangle e_n$$ and $$\Phi(a)\circ \Phi(b)(x)=\Phi(a)\left(\sum_nb_n \langle x,e_n\rangle e_n \right)=\sum_{m}a_m\left\langle \left(\sum_nb_n \langle x,e_n\rangle e_n \right),e_m\right\rangle e_m$$ For $n\neq m$ the scalar product is $0$, hence $$\sum_na_n \sum_nb_n\langle x,e_n\rangle\langle e_n,e_n\rangle e_n=\sum a_n\sum b_n\langle x,e_n\rangle e_n.$$
Can I bring the series under one sum?
The sum over $n$ consists of a single term. Since $$\left\langle \left(\sum_nb_n \langle x,e_n\rangle e_n \right),e_m\right\rangle = \sum_n b_n \langle x,e_n \rangle \langle e_n,e_m \rangle = b_m \langle x,e_m \rangle$$ you get $$\sum_{m}a_m\left\langle \left(\sum_nb_n \langle x,e_n\rangle e_n \right),e_m\right\rangle e_m = \sum_m a_m b_m \langle x,e_m \rangle e_m.$$