Does isomorphisms of sheaf of holomorphic sections implies isomorphisms of two holomorphic vector bundles over the same complex space ? My knowledge is very limited for complex geometry. I have the following question:
If we have two complex vector bundles $E\to X$ and $F\to X$ such that we have an isomorphism $\mathcal O\left(E\right) \cong \mathcal O\left(F\right)$ between the sheaf of holomorphic sections, do we have an isomorphism $E \cong F$ ?
 A: 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.
A: Yes, this is true as soon as you assume that the isomorphism of sheaves $\mathcal O(E)\cong\mathcal O(F)$ is $\mathcal O_X$-linear. 
To prove, pick a cover $(U_i)_{i\in I}$ of $X$ be opens which are trivializing both for $E$ and $F$. Now if you have an isomorphism of sheaves $\phi:\mathcal O(E)\to\mathcal O(F)$ then you can restrict it to each of these opens and get holomorphic functions $\phi_i:U_i\to GL_n(\mathbb{C})$. If you write carefully the gluing property you'll get something like: on intersections, $f_{ij}\phi_i=\phi_je_ {ij}$, where $e_{ij}$ and $f_{ij}$ are transition functions for $E$ and $F$. 
This defines a bundle map. 
A: Another way of looking at the problem is to consider the (set-valued) sheaf of isomorphisms $E\to F$ and the sheaf of isomorphisms $\mathcal O(E)\to\mathcal O(F)$. There is clearly a map from the former to the latter, so we only need to show that it is an isomorphism locally, i.e., we may assume that both $E$ and $F$ are trivial. But then, isomorphisms $E\cong F$ as well as isomorphisms $\mathcal O(E)\cong\mathcal O(F)$ both have an explicit description by elements of $GL_n(\Gamma(U,\mathcal O))$, and it is easily checked that they correspond.
