Is $U^{-1} (A \otimes_R A) \cong U^{-1} A \otimes_R U^{-1} A$? Background: EGA IV states in Proposition 16.4.14 (own translation)

Let $\rho: A \to B$ be a ring homomorphism, $S$ a multiplicatively closed subset of $B$. Then the canonical homomorphisms
  $$ S^{-1} P^n_{B/A} \to P^n_{S^{-1}B/A}$$
  arising from the canonicals homomorphisms $P^n_{B/A} \to P^n_{S^{-1}B/A}$ [...] are bijective.

Here $P^n_{B/A} = B \otimes_A B / I^{n+1}$ where $I$ is the kernel of the multiplication map $B \otimes_A B \to B, s \otimes t \mapsto st$. $P^n_{B/A}$ is regarded as an $B$-algebra via $b (\overline{s \otimes t}) = \overline{bs \otimes t}$.
Question: The Proof claims that $S^{-1} (B \otimes_A B) = (S^{-1} B) \otimes_A (S^{-1} B)$, but I don't see how this should work. 
One of the problems is that its not completely clear how the localization of the tensor is done, since S is a priori not a subset of $B \otimes_A B$.
 A: It might be a mistake, since it cites EGA I, 1.3.4 which only applies if $S$ is a multiplicative subset of $A$, not $B$. The $B$-algebra structure on $P_{B/A}^{n}$ is via the first factor, see EGA IV (16.3.7).
For an explicit example of $A,B,S$ such that $S^{-1}B \otimes_{A} B \not\simeq S^{-1}B \otimes_{A} S^{-1}B$, we can take $A = \mathbb{Z}$, $B = \mathbb{Z}[t]$, $S = \{t^{i}\}_{i \ge 0}$; then the LHS is $\mathbb{Z}[t_{1}^{\pm},t_{2}]$ whereas the RHS is $\mathbb{Z}[t_{1}^{\pm},t_{2}^{\pm}]$. They are not isomorphic because their unit groups are not isomorphic.
For the original proposition you may want to consider the following fact:

Let $f : R \to R_{0}$ be a surjective ring homomorphism whose kernel is nilpotent. If an element $r \in R$ is such that $f(r)$ is a unit of $R_{0}$, then $r$ is a unit of $R$.

Applying this to $S^{-1}P_{B/A}^{n} \to S^{−1}B$ gives you that elements of the form $1 \otimes s$ in $S^{-1}P_{B/A}^{n}=(S^{-1}B \otimes_{A} B)/S^{-1}I^{n+1}$ are units (this should imply that the desired map $S^{-1}P_{B/A}^{n} \to P_{S^{-1}B/A}^{n}$ is surjective).
(I hope you can find a better reference. You might try googling variations of "infinitesimal neighborhood of the diagonal".)
