The answer is no: In general this is not enough information to conclude $\xi$ is pure exact. This question is discussed in detail in T.Y. Lam's book "Lectures on Modules and Rings", at the end of Section 4. I recommend reading the presentation there, but I can summarize the process of obtaining a counterexample.

Start with a nonzero commutative ring $R$ for which the family $\mathcal{F}$ of all nonzero ideals has trivial intersection. Then set $A = R$ and $B=\prod_{I \in \mathcal{F}} R/I$, with $i : A \rightarrow B$ the natural map. Then $i$ is injective, and one shows that it remains injective after tensoring with any $R/I$ (4.95 in LMR). So the sequence $\xi: 0\rightarrow A \rightarrow B \rightarrow B/A \rightarrow 0$ satisfies your assumptions.

Next, one can show that $i \otimes_R N$ is not injective when $N$ is any finitely presented left $R$-module with $\bigcap_{I\in \mathcal{F}} IN \neq 0$ (4.96 in LMR).

For an explicit counterexample Lam applies the above to $R=k[x,y]/(x^2,xy,y^2)$ over a field $k$, where $N=k^3$ with $x$ and $y$ acting via $x(a,b,c) = (0,0,a)$ and $y(a,b,c) = (0,0,b)$. (Note $N$ is the indecomposable injective $R$-module.)

Finally, Lam notes that any $\xi : 0\rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ with $\xi \otimes_R R/I$ exact for all ideals $I$ *and* $B$ projective is pure exact. This is a result of Fieldhouse, which is also included as an exercise.