It's not clear whether you are working in the setting of Lie theory, or abstract group theory, or something else. This answer addresses the Lie theory aspect of the matter.

Let's focus on Lie groups whose Lie algebra is semisimple (as solvable radicals mess up things in too many ways, as usual). By serious theorems, the functor $\mathbf{G} \rightsquigarrow
\mathbf{G}(\mathbf{C})$ from connected semisimple $\mathbf{C}$-groups to connected
complex Lie groups with semisimple Lie algebra is an equivalence of categories. So all connected complex Lie groups with semisimple Lie algebra admit a unique and
functorial "linear algebraic" structure.

Let's say that a connected Lie group $G$ with semisimple Lie algebra is *linear* if $G = \mathbf{G}(\mathbf{R})^0$ for a connected semisimple $\mathbf{R}$-group $\mathbf{G}$.
From the viewpoint of "semisimple" Lie theory, the failure of this condition is a bit tricky to think about because non-isomorphic
connected semisimple $\mathbf{R}$-groups
can yield isomorphic connected Lie groups of $\mathbf{R}$-points, the most famous being the degree-$n$ isogeny ${\rm{SL}}_n \rightarrow {\rm{PGL}}_n$ over $\mathbf{R}$ with an odd $n > 1$ (this becomes an isomorphism on $\mathbf{R}$-points). Nonetheless, we can characterize it in terms of the complex-analytic theory as follows.

Consider a connected Lie group $G$ over $\mathbf{R}$ whose Lie algebra $\mathfrak{g}$ is semisimple. Dropping any semisimplicity hypotheses on Lie algebras for a moment, there is a general notion of *complexification* of $G$, namely a homomorphism $r_G:G \rightarrow G_{\mathbf{C}}$ to a complex Lie group $G_{\mathbf{C}}$ that is initial among all homomorphisms $\rho:G \rightarrow H$ to a complex Lie group (i.e., there is a unique holomorphic homomorphism $f:G_{\mathbf{C}} \rightarrow H$ such that $f \circ r_G = \rho$).
This is constructed in complete generality in Bourbaki LIE Chapter III, for example. In general ${\rm{Lie}}(G_{\mathbf{C}})$ is a quotient of $\mathfrak{g}_{\mathbf{C}}$, so
when $\mathfrak{g}$ is semisimple this quotient is semisimple and hence $G_{\mathbf{C}}$ is (canonically) linear over $\mathbf{C}$.

Obviously if $G = \mathbf{G}(\mathbf{R})^0$ for a connected semisimple $\mathbf{R}$-group
$\mathbf{G}$ then the resulting closed embedding $G \rightarrow \mathbf{G}(\mathbf{C})$ factors uniquely through a holomorphic map $G_{\mathbf{C}} \rightarrow \mathbf{G}(\mathbf{C})$ via composition with $r_G$, so $\ker r_G = 1$.

Remarkably, the converse holds:
if $\ker r_G = 1$ then $G$ is the identity component of the group $\mathbf{G}(\mathbf{R})$ of $\mathbf{R}$-points of a connected semisimple $\mathbf{R}$-group $\mathbf{G}$
(and $r_G$ is actually a closed embedding). Indeed, the canonical "algebraization" of $G_{\mathbf{C}}$ has Weil restriction $G'$ over $\mathbf{R}$ that is a connected semisimple $\mathbf{R}$-group such that $r_G$ is identified with an injective map $G \rightarrow G'(\mathbf{R})$ between connected Lie groups. In particular, $\mathfrak{g}$ is identified with a semisimple Lie subalgebra of ${\rm{Lie}}(G')$, so by the algebraic theory over $\mathbf{R}$ (as over any field of characteristic 0) it has the form
${\rm{Lie}}(\mathbf{G})$ for a unique connected semisimple closed
$\mathbf{R}$-subgroup $\mathbf{G} \subset G'$. Thus, $r_G$ factors through
$\mathbf{G}(\mathbf{R})^0$. The resulting injective map $G \rightarrow \mathbf{G}(\mathbf{R})^0$ between *connected* Lie groups is an isomorphism on Lie algebras and thus is surjective, so it is an isomorphism of Lie groups.

The upshot is that a connected Lie group $G$ with semisimple Lie algebra is linear if and only if $\ker r_G = 1$, in which case $r_G$ is a closed embedding. You may therefore think of the non-triviality of $\ker r_G$ (i.e., the absence of "enough" homomorphisms to complex Lie groups to separate points) as the exact obstruction to $G$ being linear in the sense defined above.