(\sum_n |S_n|^p)^{1/p}

(\sum_n |y_n|^q)^{1/q}

|\sum S_n(y_n)| \leq \sum |S_n| |y_n| \lw

So I think this works. Fix $1<r<\infty$ and let $r'$ be the conjugate index to $r$ so $1/r + 1/r' = 1$.

Let $$ \tau = \sum_n S_n \otimes T_n \in \mathcal{A}(\ell^q,\ell^p) \widehat\otimes \mathcal{A}(\ell^p, \ell^q), $$ where by definition, $\sum \|S_n\| \|T_n\| < \infty$. By rescaling the $S_n$ and $T_n$ we may suppose that $$ \sum_n \|S_n\|^{r'} < \infty, \quad \sum_n \|T_n\|^r < \infty. $$

Define $T:\ell^p \rightarrow \ell^r(\ell^q)$ by $T(x) = \big( T_n(x) \big)$. This makes sense as $$ \|T(x)\|^r = \sum_n \|T_n(x)\|^r \leq \|x\|^r \sum_n \|T_n\|^r \qquad (x\in \ell^p). $$ For later, note that if $N$ is large and $T'(x) = (T_1(x), T_2(x), \cdots, T_N(x), 0, 0, \cdots)$ then $\|T(x) - T'(x)\| \leq \|x\| \Big(\sum_{n>N} \|T_n\|^r\Big)^{1/r}$ and so $T'$ approximates $T$ in norm if $N$ is large.

Similarly we have $S:\ell^{p'} \rightarrow \ell^{r'}(\ell^{q'})$ given by $S(y) = \big( S_n^*(y) \big)$, using that $S_n^*: \ell^{p'} \rightarrow \ell^{q'}$ and that $\|S_n^*\| = \|S_n\|$ for each $n$. Then a simple calculation shows that $$ S^* : \ell^r(\ell^q) \rightarrow \ell^p; \quad (x_n) \mapsto \sum_n S_n(x_n). $$ Thus $S^*T= \sum_n S_n T_n = \Delta(\tau)$.

Let $r=q$ and use the natural isomorphism $\ell^q(\ell^q) \cong \ell^q$ to see that $T:\ell^p\rightarrow \ell^q$ and $S^*:\ell^q\rightarrow\ell^p$. As the isomorphism is compatible with the dual pairing, we still have $S^*T=\Delta(\tau)$.

It remains to argue that $T,S^*$ are approximable (limit of finite-rank) operators. It is clear that $T'$ (as defined above) is approximable, and so as $\mathcal A(E,F)$ is closed for any Banach spaces $E,F$, we see that $T$ is indeed approximable. Thus so is $S$, and hence $S^*$ by a well-known theorem (I don't recall who to attribute this to, but it follows from the Principle of Local Reflexivity).