...a matrix Lie group is a closed subgroup of $\mathrm{GL}_n(\mathbb{C})$.
This is correct!
There is also the notion of a Lie group...
A Lie group is a tuple $(G,\cdot, \tau,\mathcal{A})$, where $G$ is a set, $\cdot$ is a group operation on $G$, $\tau$ is a topology on $G$ which turns $G$ into a topological manifold, and $\mathcal{A}$ is a maximal smooth atlas, such that the maps
$$m\colon G\times G\to G,\quad (g,h)\mapsto g\cdot h$$
$$i\colon G\to G,\quad g\mapsto g^{-1}$$
are smooth.
What, for example, is the isomorphism? Is it of abstract groups, manifolds, or ...?
Two Lie groups are said to be isomorphic if they are isomorphic as sets, groups, topological spaces and smooth manifolds. But conveniently enough, it is actually enough to check that they are isomorphic as topological groups, as one can show that every continuous group homomorphism between Lie groups is automatically smooth.
I have also heard something saying that all Lie groups are in fact isomorphic to a matrix Lie group. Is this correct?
This is not true! A classical counterexample is $\widetilde{\mathrm{SL}_2(\mathbb{R})}$ as mentioned in other answers.
Another classical counterexample (discussed in a paper by G. Birkhoff from 1936) is obtained by taking the 3-dimensional Heisenberg group
$$G_3=\Bigg\{\begin{pmatrix}1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\end{pmatrix}:x,y,z\in\mathbb{R}\Bigg\}$$
and quotienting out by the normal subgroup
$$N=\Bigg\{\begin{pmatrix}1 & 0 & n\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}:n\in\mathbb{Z}\Bigg\}.$$
The resulting factor group
$G_3^*:=G_3/N$ can be equipped with the quotient topology. This will turn $G_3^*$ into a topological manifold, homeomorphic to the product $\mathbb{R}\times\mathbb{R}\times S^1$, which in turn is a smooth manifold whose maximal smooth atlas we can let $G_3^*$ inherit. It is easy to verify that the group structure on $G_3^*$ is "compatible" with this atlas, in the sense that both multiplication and inversion become smooth maps.
We conclude that $G_3^*$ equipped with the group operation, topology and smooth maximal atlas described above is a Lie group. However, it is not isomorphic (as Lie groups) to a matrix Lie group.
To prove this, one can show an equivalent statement, namely that $G_3^*$ doesn't admit any faithful finite-dimensional complex representations. This is done in Section 4.8 of Lie Groups, Lie Algebras and Representations by B.C. Hall. The idea behind his proof is to note that every representation $$\Sigma\colon G_3^*\to \mathrm{GL}_n(\mathbb{C})$$ of $G_3^*$ gives rise to a representation $$\Pi=\Sigma\circ \Phi\colon G_3\to \mathrm{GL}_n(\mathbb{C})$$ of $G_3$, where $\Phi\colon G_3\to G_3^*=G_3/N$ is the natural projection. By passing to the Lie algebra level, one can show (this takes some effort) that for every such representation $\Pi$, it holds that $$\ker(\Pi)\supsetneq \ker(\Phi)=N\,,$$ which implies that the representation $\Sigma$ must have a non-trivial kernel and thus cannot be faithful.
So, no, it's not true that every Lie group is isomorphic to a matrix Lie group. Interestingly, though, it actually is true for compact Lie groups:
Theorem. Every compact Lie group $G$ admits a faithful finite-dimensional representation, and is thus isomorphic to a matrix Lie group.
This is typically proved as a corollary of the famous Peter-Weyl theorem (see for instance Section IV.3 of Lie Groups Beyond an Introduction by A.W. Knapp).
