Let $\mathcal{H}$ be a Hilbert space with $\dim(\mathcal{H})=\infty$ and let $1_\mathcal{H}$ denote the identity on $\mathcal{H}$.
Then we may form the algebraic tensor product $B(\mathcal{H})\odot 1_\mathcal{H}$ (where $1_\mathcal{H}$ is an abuse of notation for the algebra $\mathbb{C}\cdot 1_\mathcal{H}$). Then the map $B(\mathcal{H})\ni x \mapsto x\otimes 1_\mathcal{H}\in B(\mathcal{H}\otimes \mathcal{H})$ should be SOT-continuous if I didn't made a mistake (this is pretty obvious on elementary tensors which are total in $\mathcal{H}\otimes \mathcal{H}$ and since $\|x\otimes 1_\mathcal{H}\|=\|x\|$, a convergent net in $B(\mathcal{H})$ gets mapped to an eventually bounded net in $B(\mathcal{H}\otimes \mathcal{H})$, hence this should suffice).
Now let $(e_n)_{n\in \mathbb{N}}$ be an orthonormal system in $\mathcal{H}$ and let $\xi=\sum_n \frac{1}{n} e_n \otimes e_n \in \mathcal{H}\otimes \mathcal{H}$. Then consider the functional $B(\mathcal{H}\otimes \mathcal{H})\ni z \mapsto \langle z\xi, \xi\rangle$. This is SOT continuous since it is induced by a rank-1 operator.
Next consider $\omega: B(\mathcal{H})\ni x\mapsto \langle x\otimes 1_\mathcal{H}(\xi),\xi\rangle$. We obtain $\omega(x)=\mathrm{tr}(\Phi x)$ with $\Phi=\sum_n \frac{1}{n^2} \langle \cdot, e_n\rangle e_n$. Since $\Phi$ is not a finite rank operator, $\omega$ cannot be SOT-continuous.
On the other hand we have $\omega(x)=x\mapsto x\otimes 1=z\mapsto \langle z\xi, \xi\rangle$, hence it is a composition of SOT-continuous maps.
Obviously I made a mistake somewhere but I cannot identify it.
What you achieved is a proof that $x\longmapsto x\otimes 1_H$ is not SOT-continuous. The fail in your argument is (as usual; we all do that) in the sentence that includes the words "pretty obvious": you say that "a convergent net in $B(H)$ gets mapped to an eventually bounded net in $B(H\otimes H)$". A SOT-convergent net is not necessarily bounded (see this question for an example of an unbounded SOT-convergent net; in this answer there is a proof of the fact that there is a net $\{\sqrt{n_j}e_{n_j}\}$ that converges weakly to zero).