Let $A$ be an $E_{\infty}$-ring, there exists the relative tensor of Bimodules (see Higher Algebra section 4.4 ); let $M$ be an right $A$-module and $N$ be a left $A$-module, the relative tensor is defined as following
$ -\otimes - :RMod_{A}\times LMod_{A}\cong\; _{\mathbb{S}}BMod_{A}\times \;_{A}BMod_{\mathbb{S}}\to\;_{\mathbb{S}} BMod_{\mathbb{S}}\cong Sp:\\ \; \; \;\; \; \; (M,N)\mapsto A\otimes B:= Bar_{A}(M,N), $
with the token $\mapsto$ I have denoted its behaviour on the objects.
Since $A$ is an $E_{\infty}$-ring, being a right $A$-module being a left $A$-module is the same, in short there is an equivalence of $\infty$-categories $RMod_{A}\cong LMod_{A}$.
Let me denote with $\hat{M}$ and with $\hat{N}$ when we consider $M$ as left $A$-Module and $N$ as right $A$-module, we can take the tensor of them $\hat{N}\otimes\hat{M}= Bar_{A}(\hat{N},\hat{M})$.
My question is the following: $M\otimes N$ and $\hat{N}\otimes\hat{M}$ are isomorphic as objects of $Sp$?
My guess is that they are isomorphic. That is why, the $n$-level of the two sided simplical bars are $ Bar_A(M,N)_n = M \otimes A^n \otimes N$ and $Bar_{A}(\hat{N},\hat{M})_n= \hat{N}\otimes A^{n}\otimes \hat{M}$ (here $\otimes$ is the smash product in $Sp$), they are isomorphic because the smash product is symmetric, hence their geometric representation is isomorphic in $Sp$.
Yes, this is true. Lurie shows in (HA, 4.5.1.4 and 4.5.1.6) that the equivalence between left and right $A$-modules looks like $\mathrm{LMod}_A\xleftarrow{U_L}\mathrm{Mod}_A\xrightarrow{U_R}\mathrm{RMod}_A$, where both $U_L$ and $U_R$ are equivalences (and the names of these arrows are mine, Lurie would have different notation). Here, $\mathrm{Mod}_A$ is basically $(A,A)$-bimodules. It is (HA, 4.5.2.1) that $\mathrm{Mod}_A$ is a symmetric monoidal $\infty$-category, so since both $U_L$ and $U_R$ are equivalences, your question is solved once we show that the diagram $$\require{AMScd} \begin{CD} \mathrm{Mod}_A\times\mathrm{Mod}_A @>{\otimes}>> & \mathrm{Mod}_A \\ @V{U_R\times U_L}VV & @VV{\mathrm{forget}}V\\ \mathrm{RMod}_A\times\mathrm{LMod}_A @>{\otimes}>>& \mathrm{Sp} \end{CD} $$ commutes. But this is true as both composites are computed by the bar construction, for instance by (HA, 4.5.2.1).