No, the multiplicities of \lambda _1 = 1 and \lambda _2 = 2 are 1. E_1 and E_2 are subspaces: they don't have multiplicities, they have dimensions. The multiplicity for each \lambda gives you the dimension of the subspaces spanned by the eigenvectors corresponding to that \lambda. Since, in general, the subspace spanned by the eigenvectors corresponding to \lambda is a subspace of E_{\lambda} (A), if the generalized eigenspace has dimension equal to the multiplicity of the eigenvalue, then it has dimension equal to the dimension of the eigenspace, hence it is the eigenspace. Since, in general, the generalized eigenvectors span the corresponding vector space, and since in this particular case, the generalized eigenvectors are the eigenvectors, the eigenvectors span the vector space. Thus, as you should know, this means that A is diagonalizable.
So to clarify:
A is diagonalizable iff the eigenvectors span the vector space
In general, the generalized eigenvectors span the vector space
Therefore, A is diagonalizable iff the eigenvectors are the generalized eigenvectors
In general, the span of the eigenvectors forms a subspace of the span of generalized eigenvectors
Therefore, the generalized eigenvectors are the eigenvectors iff the dimension of the span of the generalized eigenvectors is equal to the dimension of the span of the eigenvectors
In general, given an eigenvalue, the dimension of the span of the corresponding eigenvectors is equal to the multiplicity of that eigenvalue, (in the characeristic polynomial)
Therefore, given an eigenvalue, the span of the generalized eigenvectors equals the span of the eigenvectors iff the span of the generalized eigenvectors has dimension equal to the multiplicity of that eigenvalue.
So, where A is a square matrix, dim(E_{\lambda}(A)) equals the multiplicity of \lambda for each \lambda iff A is diagonalizable.
In your example, E1 = Span{e1} = Span{(1 0)T}. This has dimension 1, and the eigenvalue 1 has multiplicity 1 in the characteristic polynomial (x - 1)(x - 2). E2 = Span{e2} = Span{(0 1)T}. This has dimension 1, and the multiplicity of 2 is 1, so for every eigenvalue of A, the dimension of the corresponding generalized eigenspace equals the multiplicity of the eigenvalue, so from what we proved above, A is diagonalizable. Of course, look at A: it is already diagonal! We know it's diagonalizable without doing all this, but hopefully this illustrated the point.
EDIT: If you look at the second paragraph above, you can find a number of conditions that are equivalent to the diagonalizability of A. With this, it's easy to find a number of conditions that are equivalent to the non-diagonalizability of A. Find these, and hopefully that answers your question as to when A is not diagonalizable.
