correspondence between invariant forms and Lie groups

I don't have a full answer yet. Some notation: let's write $mathrm{Inv}(G)$ for the collection (algebra) of all invariant tensors for $G subseteq GL_n$, and $mathrm{Grp}(I)$ for the group of matrices that leave invariant some collection $I$ of tensors.

Then a necessary condition for $G = mathrm{Grp}(mathrm{Inv}(G))$ is for $G$ to be Zariski-closed in $GL_n$. (Recall that $GL_n$ is a codimension-one Zariski-closed subset of affine $(n^2 +1)$-space, where the first $n^2$ coordinates are the matrix coefficients, and the last one is the inverse determinant; $GL$ is cut out by the condition that the actual determinant times the last coefficient is unity.) Indeed, $mathrm{Grp}(I)$ is Zariski-closed for any $I$, because it is the intersection of $mathrm{Grp}(i)$ over all $iin I$, and fixing a tensor is a Zariski-closed condition, because $GL_n to mathrm{End}(V)$ is polynomial for $V$ and tensor product of the $n$-dimensional representation and its dual.

So this rules out things like the irrational line in the torus (diagonal two-by-two matrices with eigenvalues $exp(x)$ and $exp(pi x)$ as $x$ ranges over the field, and $pi$ is your favorite irrational number).

I think that a sufficient condition is for $G$ to be compact. This is because if you know all the tensor invariants, then you know the full subcategory of representations that are tensor-generated by the defining representations, and in fact all the subrepresentations of these, and if $G$ is compact then this category is equivalent to the full category of representations and knows the group by Tannakian arguments. But this is much too strong --- $SL_n(mathbb{R})$ is not compact, for example.