According to the comments in this recent MO question: <a href="http://mathoverflow.net/questions/59484/can-a-simple-lie-algebra-be-determined-by-weights-of-its-representation">Can A Simple Lie Algebra Be Determined By Weights of its Representation</a>, the answer to your question is no with simple examples afforded by the defining 2n-dimensional representations of Lie algebras of types $C_n$ and $D_n$.

However, unless I am overlooking some very subtle point, this result is only dependent on the geometry of the weights (i.e. the angles between the different weights within the weight lattice). If instead one looks at the set of weights of a given representation as expressed in terms of the fundamental weights then the set of weights does determine the corresponding algebra (from the set of weights one can reconstruct the Cartan Matrix and hence the algebra). 

For example, the defining 10-dimensional representation of $C_5$ (with highest weight $\omega_1$) has the following weights:

$\pm\omega_1,\pm(\omega_2-\omega_1), \pm(\omega_3-\omega_2), \pm(\omega_4-\omega_3)$, and $\pm(\omega_5-\omega_4)$

On the other hand the defining 10-dimensional representation of $D_5$ (with highest weight $\omega_1$) has the following weights:

$\pm\omega_1, \pm(\omega_2-\omega_1), \pm(\omega_3-\omega_2), \pm(\omega_4+\omega_5-\omega_3)$, and $\pm(\omega_4-\omega_5)$

In the above I am  using the conventions used by LiE for indexing the fundamental weights. When expressed in terms of the fundamental weights, one sees that the sets of weights do distinguish between $C_5$ and $D_5$ even though geometrically they are indistinguishable.