Let $(E_\lambda)_{\lambda\in L}$ be a family of right $A$-modules and $F$ a finitely generated free left $A$-module. Then the $\mathbf{Z}$-module mapping $$f:\left(\prod_{\lambda\in L}E_\lambda\right)\otimes_AF\rightarrow\prod_{\lambda\in L}(E_\lambda\otimes_AF)$$ such that $f((x_{\lambda})_{\lambda\in L}\otimes y)=(x_\lambda\otimes y)_{\lambda\in L}$ is bijective.
This map is obviously injective. I am trying to prove surjectivity. I am given the following hint:
By virtue of the fact that the $\mathbf{Z}$-module mapping $$g:\bigoplus_\lambda E_\lambda\otimes_A\bigoplus_\mu F_\mu\rightarrow\bigoplus_{\lambda,\mu}(E_\lambda\otimes_A F_\mu)$$ such that $g((x_\lambda)_\lambda\otimes(y_\mu)_\mu)=(x_\lambda\otimes y_\mu)_{\lambda,\mu}$ is bijective, this result is reduced to the case where $F=A_s$. (Note that the families $(E_\lambda) $ and $(F_\mu)$, here, are arbitrary right $A$-modules and left $A$-modules, respectively.)
Attempt:
Let $(b_\mu)_{\mu\in M}$ be a finite basis of $F$. The mapping $g:A_s^{(M)}\rightarrow F,\,\xi\mapsto\sum_\mu\xi_\mu b_\mu$, is the unique $A$-module isomorphism such that $b_\mu=g(e_\mu)$, where $(e_\mu)_\mu$ is the canonical basis of $A_s^{(M)}$.
Furthermore, the mapping $\varphi:\left(\prod_\lambda E_\lambda\right)^{(M)}\rightarrow\left(\prod_\lambda E_\lambda\right)^{(M)}\otimes_AF,\,x\mapsto\sum_\mu x_\mu\otimes b_\mu$ is a $\mathbf{Z}$-module isomorphism.
Also, for each $\lambda\in L$, the mapping $h_\lambda:E_\lambda\otimes_A A\rightarrow E_\lambda$ such that $h_\lambda(x\otimes\alpha)=x\alpha$, for $x\in E_\lambda$ and $\alpha\in A$, is an $A$-module isomorphism. Therefore $$h:\prod_\lambda(E_\lambda\otimes_A A)\rightarrow(\prod_\lambda E_\lambda),\,z\mapsto(h_\lambda(z_\lambda))_\lambda$$ is an $A$-module isomorphism.
I am not sure how to utilize the hint. Any suggestions?
Edit:
Merely assuming $F$ is free forces $f$ to be injective. It, then, seems that requiring $F$ to also be finitely generated turns $f$ into a bijection. But why does $F$ being finitely generated imply surjectivity?
Note that by definition, $A_s^{(M)}=\prod_{m\in M}A_s$ and because $M$ is finite, $\prod_{m\in M}A_s=\oplus_{m\in M}A_s$. Then by the hint, there is a $\mathbb{Z}$-module isomorphism
$$\phi:\left(\prod_{\lambda\in L} E_\lambda\right)\otimes_A \left(\bigoplus_{m\in M}A_s\right)\to\bigoplus_{m\in M}\left(\left(\prod_{\lambda\in L}E_\lambda\right)\otimes_A A_s\right).$$
Combine this with your other observations to reduce the problem to the case of establishing that the map
$$f_0:\left(\prod_{\lambda\in L}E_\lambda\right)\otimes_A A_s\to\prod_{\lambda\in L}\left(E_\lambda\otimes_A A_s\right)$$
is an isomorphism.