Let $i : U \to X$ be a quasi-compact open immersion of schemes. Under which conditions is the natural map
$i_* M \otimes i_* N \to i_* (M \otimes N)$
for all $M,N \in \text{Qcoh}(U)$ an isomorphism? We may assume that $X=\text{Spec}(A)$ is affine. If $U$ is affine, then we may reduce to the case $M=N=\mathcal{O}_U$ (using presentations) and use that $i^{\#}$ is a flat epimiorphism to get an affirmative answer. If there are counterexamples for general $U$, what conditions are sufficient?
Here's a counterexample: over a field $k$, let $X=\mathbb{A}^4=\mathbb{A}^2\times \mathbb{A}^2$, and $U$ the complement of the origin. Put $Y=\mathbb{A}^2\times\{0\}$, $Z=\{0\}\times\mathbb{A}^2$, $Z'=Z\cap U$, $Y'=Y\cap U$.
Take $M=\mathcal{O}_{Y'}$ and $N=\mathcal{O}_{Z'}$. Then $M\otimes N$ is zero, while $ i_* M=\mathcal{O}_{Y}$ and $i_* N=\mathcal{O}_{Z}$, so $i_* M\otimes i_* N$ is the structure sheaf of the origin.
This is an addition to the Laurent's answer. First, it should be said that if one derives all the functors, one will get an isomorphim --- $Ri_*M \otimes^L Ri_*N \cong Ri_* (M \otimes^L N)$. Indeed, it is a simple corollary of the projection formula: $$ Ri_*M \otimes^L Ri_*N \cong Ri_*(M \otimes^L i^*Ri_*N) \cong Ri_*(M \otimes^L N) $$ (the second isomorphism is by the flat base change). What goes wrong with the underived version is that $M$ and $N$ have higher direct images which then have $Tor$'s all of which eventually get canceled. In the particular example of Laurent one has $$ R^1i_*O_{Y'} = y_1^{-1}y_2^{-1}k[y_1^{-1},y_2^{-1}], \quad R^1i_*O_{Z'} = z_1^{-1}z_2^{-1}k[z_1^{-1},z_2^{-1}], $$ where are $y_1,y_2$ are coordinates on $Y$ and $z_1,z_2$ are coordinates on $Z$. In addition to $O_Y\otimes O_Z = k$ we have $$ Tor_2(O_Y,R^1i_*O_{Z'}) = Tor_2(R^1i_*O_{Y'},O_Z) = k, $$ $$ Tor_4(R^1i_*O_{Y'},R^1i_*O_{Z'}) = k, $$ and it is easy to see that all this cancels in the spectral sequence calculating $Ri_*O_{Y'} \otimes^L Ri_*O_{Z'}$.
