Suppose $f:X\to Y$ is a morphism of smooth connected schemes (over some base). Say $Z\subseteq Y$ is a closed subscheme with complement $U$ so that $f$ pulls back (restricts) to isomorphisms on $Z$ and $U$. Does it follow that $f$ is an isomorphism?
If we drop the condition that $X$ and $Y$ have to be smooth, the normalization of a cusp is a counterexample. If we remove the hypothesis that $X$ and $Y$ must be connected, taking $X=U\sqcup Z$ gives a counterexample.
Reducing to the affine case, the question is this:
Given a ring homomorphism $R\rightarrow S$, and given an ideal $I\subset R$, suppose that all of the following are isomorphisms: $$R/I\rightarrow S/IS$$ $$R_f\rightarrow S_f\quad \hbox{for any $f\in I$}$$ Can we conclude that $R\rightarrow S$ is an isomorphism?
Assuming irreducibility (i.e. assuming $R$ and $S$ are domains) the answer is certainly yes if $I=(f)$ is principal. Then it's easy to see that $R\rightarrow S$ is injective (because the injection $R\rightarrow R_f\approx S_f$ factors through it). For surjectivity, let $s\in S$. Then because $R_f\approx S_f$, we can write $f^ks=r$ for some $r\in R$. Then $r$ maps to zero in $S/fS$, so $r\in fR$, so (because $f$ is not a zero-divisor) $f^{k-1}s\in R$, contradicting the fact that we could have chosen $k$ minimal.
More generally, if $I=(f_1,\ldots,f_k)$ is finitely generated, then $f_i^N s\in R$ and therefore $f_i^N s\in I\subset R$ for every $i$, which at the very least forces $s$ to be integral over $R$.
I expect you can settle the general case either by generalizing this argument or by constructing $I$ and $S$ in a way that forces it to go wrong.
This is essentially the argument in Bhargav's comment. Matt Satriano showed me the separatedness argument.
We first apply the valuative criterion for separatedness to show that $f:X\to Y$ is separated. In fact, this shows that any morphism which is injective on topological spaces must be separated. Suppose $\Delta$ is the spectrum of a valuation ring with closed point $\ast$. Given any map $\Delta\to Y$, there is a unique set theoretic lift to $X$. Therefore, any failure of the valuative criterion will be witnessed on the induced map from an affine open neighborhood of $\ast$ in $X$ to an affine open neighborhood of $\ast$ in $Y$. Since any morphism of affine schemes is separated, the valuative criterion holds, so $f$ is separated.
Since $f$ induces isomorphisms on geometric points, it is quasi-finite. Supposing $Y$ is normal, integral and locally noetherian and $f$ is finite type, Zariski's Main Theorem† tells us that $f$ is an open immersion. Since $f$ is surjective, it is an isomorphism.
† I'm always surprised at how everything seems to be ZMT, so here's a precise reference: EGA III, Corollary 4.4.9. If $Y$ is normal, integral, and locally noetherian, $f:X\to Y$ is separated, birational, finite type, and quasi-finite, then $f$ is an open immersion.
