Here is a different view on your original question whether the canonical morphism $$ \tag{$\star$}
i_* \mathscr{O}_U \otimes_{\mathscr{O}_X} i_* \mathscr{O}_U \to i_* \mathscr{O}_U$$ is an isomorphism.

It seems that the question is very different depending on the dimension of the complement of $U$. Let $Z=X\setminus U$ and observe that we may break up the embedding $U\to X$ into intermediate embeddings and so assume that $Z$ is irreducible. Then we will distinguish two cases, whether $Z$ is a divisor or has higher codimension.

## 1

Assume that $\mathrm{codim}(Z,X)\geq 2$.

*Claim*: If $X$ is noetherian and $U$ is generically reduced (reduced at all the generic points), then $i_*\mathscr O_U$ is a coherent $\mathscr O_X$-module.

*Proof*: Consider the normalization $\nu:X'\to X$ and let $i':U'=\nu^{-1}U\to X'$. Then we have an injective natural map:
$$
i_*\mathscr O_U \to i_*\nu_*\mathscr O_{U'}\simeq \nu_*i'_*\mathscr O_{U'}\simeq \nu_*\mathscr O_{X'}.
$$
Since $\nu$ is proper (it is finite), $\nu_*\mathscr O_{X'}$ is a coherent $\mathscr O_X$-module and this exhibits $i_*\mathscr O_{U}$ as a submodule. $\square$

Let $p\in X$ be a point with residue field $\kappa(p)$ and consider the reduction of $(\star)$ at $p$:
$$
i_* \mathscr{O}_U \otimes_{\mathscr{O}_X} i_* \mathscr{O}_U\otimes\kappa(p) \to i_* \mathscr{O}_U\otimes\kappa(p)$$
For simplicity let's introduce the notation $V_p=i_*\mathscr O_U\otimes\kappa(p)$ so the above morphism is essentially
$$ \tag{$\star\star$}V_p\otimes_{\kappa(p)}V_p\to V_p$$

It is clear that if $(\star)$ is an isomorphism, then so is $(\star\star)$.
By the above *Claim* $V_p$ is finite dimensional and hence for $(\star\star)$ to be an isomorphism it is necessary that $\dim_{\kappa(p)}V_p=1$.

In the example of the two planes meeting in a single point this fails.

On the other hand if $X$ is reduced, then $(\star\star)$ being an isomorphism is actually equivalent to $(\star)$ being an isomorphism near $p$. Indeed, if $\dim_{\kappa(p)} V_p=1$ then this holds in a neighbourhood of $p$ and then $i_*\mathscr O_U$ is a line bundle (in the same neighbourhood) and therefore $(\star)$ is an isomorphism (in the same neighbourhood).

So (at least if $X$ is reduced and noetherian) then the remaining question is when $i_*\mathscr O_U$ is a line bundle. This happens for example if $X$ is $S_2$. In this case the natural map $\mathscr O_X\to i_*\mathscr O_U$ is an isomorphism, so the latter is trivially a line bundle. This happens for example if $X$ is normal.

In fact, being a line bundle means being locally isomorphic to $\mathscr O_X$ so this condition is actually equivalent to being $S_2$ (assuming $X$ is reduced). (To see why, consider the natural map $\mathscr O_X\to i_*\mathscr O_U$, and observe that any section of $\mathscr O_X$ would vanish on a codimension $1$ subset, and hence this map induces an embedding on the residue fields $\kappa(p)\to V_p$, but if it is an embedding, then it is an isomorphism and hence surjective. Then by Nakayama's lemma the original map is surjective on stalks and hence an isomorphism).

If we do not require $X$ to be reduced, then a little less is enough. We always have an exact sequence of sheaves:
$$
\mathscr H^0_Z(X,\mathscr O_X) \to \mathscr O_X\to i_*\mathscr O_U \to \mathscr H^1_Z(X,\mathscr O_X).
$$
This shows that as soon as $\mathscr H^1_Z(X,\mathscr O_X)=0$, it follows that $\mathscr O_X\to i_*\mathscr O_U$ is surjective. Of course, if $X$ is reduced, then $\mathscr H^0_Z(X,\mathscr O_X)=0$, so we're back at assuming that $X$ is $S_2$. I think this gives a pretty good description of what happens in case $\mathrm{codim}(Z,X)\geq 2$.

## 2

Assume that $Z\subset X$ is a divisor.

## 2a)

If $Z$ is a $\mathbb Q$-Cartier divisor, then $U$ is locally equal to the basic open set corresponding to the defining equation of $mZ$ (where $mZ$ is a multiple which is Cartier) and hence the natural map $\mathscr O_X\to i_*\mathscr O_U$ is an epimorphism since it is a localization.

## 2b)

If $Z$ is a non-$\mathbb Q$-Cartier divisor, then it seems a little more complicated, but it seems to me that it still might work.

Perhaps one can do this: blow up $X$ along $Z$ and get $\sigma: Y\to X$. This makes the preimage of $Z$ a Cartier divisor whose complement is still $U$, say $j:U\to Y$ is the embedding. Then $\mathscr O_Y\to j_*\mathscr O_U$ is an epimorphism by part 2a). Now it seems to me that the push-forward of this to $X$ remains an epimorphism since this happens on the sections on open sets and not just on the stalks.

Then if we assume that $X$ is normal, then $\mathscr O_X\to \sigma_*\mathscr O_Y$ is an isomorphism, so the desired condition follows. I am not entirely sure about this last part, but I have already written too much, so I will just leave this as is for now.