This result is true and not easy. You can find this result stated and a proof of "complexification of compact group implies reductive" in chapter 5 of these notes.
I don't know a proof of the converse that doesn't already establish a substantial part of the classification of reductive groups. In the case that $G$ is centerless and simple, you can see a proof as Lemma 2 here.
One sign that it is hard is that you need to use the hypothesis that your complex group is a linear algebraic group. For example, let $E$ be an elliptic curve over $\mathbb{C}$. Then $E$ is a group object in the category of $\mathbb{C}$ varieties which is not the complexification of any compact group. Indeed, if the $j$-invariant of $E$ is not real, then $E$ doesn't even have any anti-holomorphic involutions.
The question is quite muddled and is based on a misconception. In the theory of Lie groups, a "torus" (as a subgroup) need not be homeomorphic to a topological torus.
Let me work with complex semisimple Lie groups $G$, since the discussion is cleaner in this case and this is what the question was about. I will assume that $G$ has finitely many connected components. Then $G$ has
complex algebraic structure meaning that $G$ embeds as a Zariski closed subgroup of $GL(N, {\mathbb C})$ for some $N$ (Zariski closed means a subgroup defined via a system of polynomial equations), see here.
A torus in $G$ is a connected abelian complex Lie subgroup $H< G$ whose (almost faithful) linear representation is diagonalizable. Which representation you take, does not matter, you can take the above representation $G\to GL(N, {\mathbb C})$ or you can take the adjoint representation of $G$. (Almost faithful means that the kernel is finite.) A maximal torus (usually denoted $T$ or $A$) is a maximal (with respect to inclusion) subgroup with this property. All maximal tori are conjugate to each other. (Their Lie algebras are Cartan subalgebras, i.e. commutative self-normalizing subalgebras of the Lie algebra of $G$.)
Since $H$ is diagonalizable, it is isomorphic to a closed subgroup of $({\mathbb C}^\times)^N$, which implies that $H\cong ({\mathbb C}^\times)^n$ for some $n$.
Conversely, suppose that $H< G$ is a subgroup isomorphic to $({\mathbb C}^\times)^n$. I claim that $H$ is diagonalizable as a subgroup of $GL(N, {\mathbb C})$. Prove this by induction on $N$. The case $N=1$ is clear.
Then verify that every connected abelian subgroup of $GL(N, {\mathbb C})$ has an invariant line $L\subset {\mathbb C}^N$ (actually, this is even true for solvable subgroups). How do we know this? Since we are working over the algebraically closed field, each nontrivial element of $H$ has an eigenvector in ${\mathbb C}^N$. By commutativity of $H$, the eigenspace decomposition is $H$-invariant. Now, use the induction hypothesis. In any case, we got an invariant line, so we have another linear representation of $H$ on the vector space $V={\mathbb C}^N/L$. We need to prove that the sequence of $H$-modules splits:
$$
0\to L\to {\mathbb C}^N \to V\to 0.
$$
So far our proof used only the fact that $H$ is abelian. Now, I will use that $H$ contains the "compact subtorus" $K_H=(S^1)^n$. The next part of the proof is called the "unitary trick" (more precisely, this is a very special case of the unitary trick). Since the subtorus $K_T$ is compact, it preserves some hermitian inner product on ${\mathbb C}^N$. Thus, $K_T$ preserves $L^\perp$, the orthogonal complement to $L$ in ${\mathbb C}^N$ defined via this inner product. But $H$ is the complexification of $K_T$, hence, the entire $H$ preserves $L^\perp$. (This is a pleasant exercise in complex analysis: If $f$ is a holomorphic function on $H$ vanishing on $K_H$, then $f$ is identically zero. Now, convert this into the statement about invariance of $L^\perp$.) As a $K$-module, $L^\perp$ is isomorphic to $V$ of course. Therefore, the above sequence splits. By the induction hypothesis, $V$ is diagonalizable as an $H$-module (i.e. we have an isomorphism of $H$-modules $V\cong L_1\oplus ... \oplus L_{n-1}$ where each $L_i$ is 1-dimensional) therefore, adding the extra factor $L$ to this decomposition we obtain a diagonalization of $H$. qed
Few more remarks in the non-anlegbraically closed case. A split torus (over some field) in the language of algebraic groups, is a group isomorphic to $(G_m)^n$ for some $n$. Thus, if we treat $H=({\mathbb C}^\times)^n$ as a real algebraic group, then the split torus in $H$ is its set of real points, i.e. the subgroup $({\mathbb R}^\times)^n$. The complementary subgroup $(S^1)^n$ is sometimes called a "compact torus". This part indeed is a torus in the topological sense. But this torus will never be maximal in this situation. One more thing to note: Start with a compact Lie group $K$ (without abelian factors). Then tori in $K$ indeed are all of the form $T=(S^1)^n$. Once you complexify $K$, you obtain a complex semisimple Lie group $G$. The tori in $G$ will be (up to conjugation) of the form $T^{\mathbb C}$, complexifications of tori in $K$.
Best Answer
You already have constructed $G_0$: $\Bbb R[x,y]/(x^2+y^2+1)\otimes_{\Bbb R} \Bbb C = \Bbb C[x,y]/(x^2+y^2+1)$, and under the change of variables $u=-x-iy,v=x-iy$ we have that $\Bbb C[x,y]/(x^2+y^2+1) \cong \Bbb C[u,v]/(uv-1)$, which is of course isomorphic to $\Bbb G_m$ over $\Bbb C$.