Is every Zariski-closed real matrix Lie group the joint stabilizer of some list of mixed tensors? Consider a finite-dimensional real vector space $V$,
and an embedded real Lie subgroup $G \subset \mathrm{GL}_\mathbb{R}(V)$. In what follows, $V^*$ denotes the real dual vector space of $V$.
Def: Let us say $G$ has property $\mathcal{P}$
if there exist some finitely many $\alpha_1,\ldots,\alpha_k$ in the mixed real tensor algebra $TV$ (including tensor factors of both $V$ and $V^*$) of $V$,
such that $G= \bigcap_{j=1}^{k}\mathrm{Stab}_{\mathrm{GL}_{\mathbb{R}}(V)}(\alpha_j)$.
Here we use the induced action of $\mathrm{GL}(V)$ on the mixed tensor algebra $TV$.
My question is whether every Zariski-closed Lie subgroup of $\mathrm{GL}_{\mathbb{R}}(V)$ has property $\mathcal{P}$?
(Here we use the Zariski topology on $\mathrm{End}_{\mathbb{R}}(V)$ coming from real polynomials, e.g. from a choice of basis, or just in the multilinear sense.)

Examples having property $\mathcal{P}$:
For $V = \mathbb{R}^n$, for $G = \mathrm{O}(n,\mathbb{R})$,
we can use $\alpha \in V^* \otimes V^*$ given by the standard real-bilinear inner product.
For $V = \mathbb{R}^n$, for $G = \mathrm{SL}(n,\mathbb{R})$,
we can use
$\alpha \in V^{\otimes n}$ the standard volume form.
For $V = \mathbb{R}^{2n}$, for $G = \mathrm{GL}(n,\mathbb{C}) \subset \mathrm{GL}(2n,\mathbb{R})$,
we can use $\alpha = J = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)  \in V \otimes_\mathbb{R} V^*$.
For $V = \mathbb{R}^{2n}$, for $G = U(n)$,
if I'm not mistaken, we can use both $\alpha = J = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)  \in V \otimes_\mathbb{R} V^*$,
and the
$\beta_1,\beta_2 \in V^* \otimes_\mathbb{R} V^*
$
which correspond to the real and imaginary parts of the standard Hermitian inner product $\mathbb{C}^{n} \times \mathbb{C}^{n} \rightarrow \mathbb{C}$ but treated as a real-bilinear map.
For $V = \mathbb{R}^n$ and $G = \mathbb{R}^\times 1 \subset \mathrm{GL}(n,\mathbb{R})$
the nonzero multiples of the identity,
we should be able to take any $\alpha_1,\ldots,\alpha_{n^2} \in V \otimes_\mathbb{R} V^* \simeq \mathrm{M}(n,\mathbb{R})$
such that the $\alpha$'s form an $\mathbb{R}$-linear basis of $\mathrm{M}(n,\mathbb{R})$. (This uses the fact that only the scalar multiples of the identity commute with all other square matrices.)

Lemma: If $G$ has property $\mathcal{P}$, then $G \subset \mathrm{GL}_{\mathbb{R}}(V)$ is Zariski-closed;
i.e. it is the intersection of $\mathrm{GL}_{\mathbb{R}}(V)$
with a Zariski-closed subset (a real polynomial-vanishing subset) of $\mathrm{End}_{\mathbb{R}}(V)$.
Proof: After picking bases, each "stabilizing equation" $g \cdot \alpha_j = \alpha_j$ can be converted into a polynomial equation in the entries of $g$ and $g^{-1}$, and the $g^{-1}$ factors can be multiplied out to get a polynomial in just $g$'s entries;
thus $G$ is the vanishing set in $\mathrm{GL}_{\mathbb{R}}(V)$
of a collection of real polynomials of $\mathrm{End}_{\mathbb{R}}(V)$.
Rmk:
For instance, as pointed out in the comments below, $\mathrm{GL}_+(n,\mathbb{R})$ is not Zariski-closed in $\mathrm{GL}_(n,\mathbb{R})$, hence $\mathrm{GL}_+(n,\mathbb{R})$ cannot have property $\mathcal{P}$.

My question is: does every Zariski-closed, real Lie subgroup of $\mathrm{GL}_{\mathbb{R}}(V)$ have property $\mathcal{P}$?
(I've updated the question to reflect the suggestions noticing that every $G$ having property $\mathcal{P}$ is Zariski-closed; now I'm wondering about the converse.)
 A: As mentioned in the comments, it is not true that every Zariski-closed subgroup $H$ of $G=\mathrm{GL}_n(\mathbf{R})$ is the intersection of the stabilizers of some tensors: it fails for $n=2$ and $H$ the group of all upper-triangular invertible $2$ by $2$ matrices (the Borel subgroup).
Proof: if $H$ were the stabilizer of a finite collection of tensors, then by taking $E$ to be the appropriate direct sum of tensor spaces we would obtain $H$ as the stabilizer of a certain vector $e \in E$, where $E$ is a finite-dimensional $G$-module. Tensoring up to $\mathbf{C}$ would then imply that the group $H_\mathbf{C}$ of upper triangular invertible $2$ by $2$ matrices with entries in $\mathbf{C}$ is the stabilizer in $\mathrm{GL}_2(\mathbf{C})$ of $e \in E_\mathbf{C}=\mathbf{C} \otimes_{\mathbf{R}} E$.
But this would imply that the projective variety
$$\mathbf{P}^1(\mathbf{C}) \cong G_\mathbf{C} / H_\mathbf{C} \cong G_\mathbf{C} \cdot e \hookrightarrow E$$ is a (locally closed) sub-variety of the affine variety $E$, contradiction. (See Proposition 6.7 of Borel, Linear algebraic groups; the separability hypothesis there is automatic in characteristic $0$).
What is true is that given any algebraically closed field $K$, any subfield $k \leq K$, any algebraic $K$-group $G$ defined over $k$ and any Zariski-closed subgroup $H$ of $G$ also defined over $k$, there is a faithful representation $E$ of $G$ defined over $k$ and a one-dimensional subspace $L \subseteq E$ defined over $k$ such that
$$H=\{g \in G \ | \ h(L)=L \}.$$ This is Theorem 5.1 of loc. cit.
Since any faithful representation of $\mathrm{GL}_n(\mathbf{C})$ is realized, up to twisting its irreducible constituents by the appropriate powers of the determinant, as a subrepresentation of some $(\mathbf{C}^n)^{\otimes d}$, this shows that you can always find some line in some tensor space so that a given Zariski-closed subgroup defined over $\mathbf{R}$ is its set-wise stabilizer.
Moreover, it seems likely to me that if the subgroup is reductive, as in all your examples, then you can obtain it as the stabilizer of a vector in some tensor space: find a line as above in some tensor space $E$, and then consider $v \otimes v^* \in E \otimes E^*$, where $v$ spans the given line and $v^*$ spans an $H$-stable complement to its annihilator.
