The answer is no - the point is that finitely generated projective modules are locally free but not necessarily globally so.

For instance take a Dedekind domain $A$ which does not have unique factorization and consider a non-principal prime ideal $P$. Then $\tilde{P}_x \cong \mathcal{O}_x$ for any $x\in Spec A$ but it is not isomorphic to $A$.

This will occur for any scheme with non-trivial Picard group, in the sense that there will be line bundles (i.e. locally free sheaves of rank 1) which are not trivial.

Although if you ask that $Spec A$ be Hausdorff you probably do get the result - this is because a Hausdorff scheme is necessarily 0 dimensional. The correct notion corresponding to Hausdorff is separated and all affine schemes are separated so adding this doesn't help.

The answer is no - the point is that finitely generated projective modules are locally free but not necessarily globally so.

For instance take a Dedekind domain $A$ which does not have unique factorization and consider a non-principal prime ideal $P$. Then $\tilde{P}_x \cong \mathcal{O}_x$ for any $x\in Spec A$ but it is not isomorphic to $A$.

This will occur for any scheme with non-trivial Picard group, in the sense that there will be line bundles (i.e. locally free sheaves of rank 1) which are not trivial.

Edit If you really do want $Spec A$ Hausdorff then off the top of my head one can say the following. If $A$ is noetherian then since it must be dimension 0 it is artinian and so a product of artin local rings. So the only dense subset in the spectrum is the whole thing and so any line bundle is trivial as it is a product of line bundles over local affine schemes.