Alright to first clear up how maps of the form $(x\otimes y)\tilde\otimes(z\otimes w)$ for arbitrary $x,y,w,z\in\mathbb H$ act on general $\varphi\in S_2(\mathbb H)$, it goes like
$$
[(x\otimes y)\tilde\otimes(z\otimes w)](\varphi):=\langle \varphi,z\otimes w\rangle_\text{HS} (x\otimes y)
$$
as defined by OP in another question. One readily verifies $\langle \varphi,z\otimes w\rangle_\text{HS} =\langle \varphi(w),z\rangle_{\mathbb H}$. The above map admits the trace
$$
\operatorname{tr}[(x\otimes y)\tilde\otimes(z\otimes w)]=\sum_{k,l\geq 1}\langle [(x\otimes y)\tilde\otimes(z\otimes w)](e_k\otimes e_l),e_k\otimes e_l\rangle_\text{HS}=\sum_{k,l\geq 1} \langle (e_k\otimes e_l)(w),z\rangle_{\mathbb H}\langle x,(e_k\otimes e_l)(y)\rangle_{\mathbb H}=\sum_{k,l\geq 1} \langle w,e_l\rangle_{\mathbb H}\langle e_k,z\rangle_{\mathbb H}\langle x,e_k\rangle_{\mathbb H}\langle e_l,y\rangle_{\mathbb H}=\langle w,y\rangle_{\mathbb H}\langle x,z\rangle_{\mathbb H}
$$
by basis expansion. On the other hand, $\operatorname{tr}(a\otimes b)=\langle a,b\rangle_{\mathbb H}$ for all $a,b\in\mathbb H$ (can also be easily seen by basis expansion). Piecing things together, one gets the somewhat peculiar identity
$$
\boxed{\operatorname{tr}[(x\otimes y)\tilde\otimes(z\otimes w)]=\operatorname{tr}(x\otimes z)\operatorname{tr}(w\otimes y)}
$$
so the trace identity in your answer is not correct. This is solely due to the above definition of the tensor product $\tilde\otimes$ to go from $S_2(\mathbb H)$ to operators on $S_2(\mathbb H)$ which does not coincide with the usual tensor product (as in that case one would in fact have $\operatorname{tr}(A\tilde\otimes B)=\operatorname{tr}(A)\operatorname{tr}(B)$ given $A,B$ are trace-class). To see how $(x\otimes y)\tilde\otimes (x\otimes y)$ should behave on $\mathbb H\otimes\mathbb H$ to mimic the usual tensor product of linear Hilbert space operators, check my old answer below.
OLD ANSWER:
To identify the Hilbert-Schmidt-Operators with $\mathbb H\otimes\mathbb H$ (as Peter Melech pointed out in his comment) and to not raise confusion regarding the actual Hilbert space tensor product $\otimes_{\mathbb H}=:\otimes$, I will "rename" your map $x\otimes y$ to $\langle \cdot,y\rangle x:\mathbb H\to\mathbb H$ (but the action is the same!). Consider the map
$$
A=(\langle \cdot,y\rangle x)\otimes(\langle \cdot,y\rangle x):\mathbb H\otimes \mathbb H\to\mathbb H\otimes \mathbb H
$$
defined via
$$
a\otimes b\mapsto (\langle a,y\rangle x)\otimes(\langle b,y\rangle x)=\langle a,y\rangle\langle b,y\rangle\, x\otimes x=\langle a\otimes b,y\otimes y\rangle\, x\otimes x
$$
and its linear extension onto all of $\mathbb H\otimes\mathbb H$, i.e., $$\boxed{A(z)=\langle z,y\otimes y\rangle x\otimes x}$$ for all $z\in\mathbb H\otimes\mathbb H$. [Edit: Thus in your notation and given your definition, the identity $$(x\otimes y)\tilde\otimes(x\otimes y)=(x\,\tilde\otimes \,x)\otimes(y\,\tilde\otimes \,y)$$ has to hold to mimic the usual tensor product of linear operators.] Now on the one hand,
$$
\operatorname{tr}(A)=\operatorname{tr}(\langle \cdot,y\otimes y\rangle\, x\otimes x)=\langle x\otimes x,y\otimes y\rangle=\langle x,y\rangle^2\,.
$$
On the other hand, let $\lbrace e_k\rbrace_{k\geq 1}$ be an orthonormal basis of $\mathbb H$, then $\lbrace e_k\otimes e_l\rbrace_{k,l\geq 1}$ is an orthonormal basis of $\mathbb H\otimes\mathbb H$. By definition of the trace
$$
\operatorname{tr}(A)=\sum_{k,l\geq 1}\langle A(e_k\otimes e_l),e_k\otimes e_l\rangle=
\sum_{k,l\geq 1}\langle e_k\otimes e_l,y\otimes y\rangle\langle x\otimes x,e_k\otimes e_l\rangle\\
=\Big\langle x\otimes x, \sum_{k,l\geq 1}\langle y\otimes y,e_k\otimes e_l\rangle e_k\otimes e_l\Big\rangle=\langle x\otimes x,y\otimes y\rangle=\langle x,y\rangle^2
$$
where in the last step we just used the basis expansion in $\mathbb H\otimes\mathbb H$.
