No. On $\mathbb P^1=\mathbb P^1(\mathbb C)$ we have $\Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(-1))=\Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2)=0$, but $O_{\mathbb P^1}(-1)$ and $O_{\mathbb P^1}(-2)$ are not isomorphic.

However on an *affine algebraic* variety $X$, the answer is "yes". There is an amazing equivalence of categories between $\mathcal O(X)$-modules and the so-called quasi coherent sheaves on $X$. It is denoted $M\mapsto \tilde M.$

In particular if you have a vector bundle $E$ on $X$, you can recover it (or rather its locally free associated sheaf) from $M=\Gamma(X,E)$ by this equivalence.

And this remarkable result is not even very difficult! ( Hartshorne, *Algebraic Geometry*, II Corollary 5.5) . And it is valid on any affine *scheme*!

**Another interpretation**

I have interpreted $\mathcal O(E)$ as the vector space $ \Gamma(X,E)$ of global sections of the bundle $E$.

However Donu Arapura and Qfwfq consider that the notation designates the sheaf of sections of the vector bundle $E$, that is the sheaf $\mathcal E$ associating to the open subset $U\subset X$ the vector space $\mathcal E(U)=\Gamma(U,E)$.

In that case the answer is : yes, that sheaf determines the bundle.

Indeed there is a *canonical* way to obtain from the sheaf $\mathcal E=\mathcal O(E)$ a holomorphic vector bundle $ Vec(\mathcal E) $ isomorphic to $E$.

Its fiber at $x\in X$ is the the $\mathcal O_{X,x}/ \mathfrak m_x=\mathbb C$- vector space $Vec(\mathcal E) [x]=\mathcal E _x/\mathfrak m_x \mathcal E_x$.

And the complex structure on $Vec(\mathcal E)=\bigsqcup Vec(\mathcal E) [x]$, is obtained from bijections with $U\times \mathbb C^r$ for all $U$'s on which $\mathcal E|U$ is free, that is isomorphic to $\mathcal O^r_U$.

Of course, there are verifications to be made, which are as straightforward as they are boring and unpleasant to write down explicitly...

**Edit** This latter interpretation is the one chritian had in mind, as he just stated in a comment.