In "Cluster algebras of finite type and positive symmetrizable matrices" by Barot, Geiss and Zelevinsky we can see that positive definite quasi-Cartan matrices (defined the same as Cartan matrices but relaxing the condition of non-positivity of elements off the diagonal) actually come from positive definite Cartan matrices. That is, each positive definite quasi-Cartan matrix is equivalent to a positive definite Cartan matrix, where equivalence in the symmetric case is defined as $A\sim B$ if there is a matrix $E$ with determinant $\pm 1$ so that $A=E^TBE$. This is proved in proposition 2.9.
This means that the graphs you are looking for are classified up to the equivalence relation above. Since a such a matrix $E$ as above can be constructed easily for the case of trees for example this means that all trees that answer your question are also ADE.
A classification result that includes the trees example above is given in "Sincere weakly positive unit quadratic forms" by M.V. Zeldich (Canadian Mathematical Society, Conference proceedings, Vol 14, 1993) and again, this classification contains graphs that are obtained from ADE by some modifications, and the resulting graphs are described with 2-colored edges.
Now to the actual question, I could observe the following lemmas which seem to reduce the problem to verifying positivity for just a few types of graphs.
Lemma 1 The (bicolored, say blue for $-1$ weights and red for $+1$ weights) graph corresponding to the matrix A doesn't have any vertex of degree $4$. (The degree here refers to the sum of degrees in each colored subgraph.)
Lemma 2 If there are two vertices $u,v$ in this graph both of degree $3$, then they share more than one common neighbor.
Lemma 3 The bicolored cycles whose corresponding matrix is positive definite are the ones with an odd number of red edges.
These are trivial to prove on their own but they reduce the possible graphs to disjoint unions of the cycles described above, Dynkin diagrams and three extra types of graphs.