Let's consider the last version of the question over $\mathbb C$. We will prove two facts: **I**, **II**.

**I.** If two complex algebraic groups $G_1$ and $G_2$ are diffeomorphic, then they are either both reductive or both non reductive.

**II.** For any reductive complex algebraic group there exist at most finite number of diffeomorphic to it complex algebraic reductive groups.

*Obvious remark.* Of course, two complex Lie groups that the same as algebraic varieties are diffeomorphic.

*Proof of* **I**. This follows from the following claim.

*Claim.* A complex algebraic $n$-dimensional group is reductive if and only if it is homotopy equivalent to an (orientable) real compact manifold of dimension $n$.

Clearly, this shows that a non-reductive group can not be even diffeomorphic to a reductive one.

Let's see how to prove this claim. We will use two alternative characterisations of reductive groups.

*First*, a complex algebraic group is reductive if and only if it has a compact real form, see for example here:

http://www.math.uchicago.edu/~mbergeron/ComplexReductive.pdf

If follows from this, that any complex reductive group is homotopy equivalent to a compact manifold of half dimension, its real compact form.
See, for example, "Other characterisations" here: https://en.wikipedia.org/wiki/Reductive_group .
For example $(\mathbb C^*)^n$ is homotopy equivalent to $(S^1)^n$. In particular $H_n(G,\mathbb Z)$ of a complex reductive group $G$ of dimension $n$ is $\mathbb Z$.

*Second.* The mid-dimensional homology group vanishes for non-reductive groups, even more, non-reductive groups are homotopy equivalent to compact manifolds of dimensions less than the half. To prove this one uses the following definition of reductive groups: A non-reductive group has a nontrivial normal unipotent subgroup.
See here https://en.wikipedia.org/wiki/Reductive_group.
Now, if we quotient a group by a normal unipotent subgroup, the homotopy type doesn't change. So we can quotient until we get a reductive group.

*Proof of* **II.** If is easy to see from the proof of **I**, that if $G_1$ is diffeomorphic to $G_2$, then their (half-dimensional) compact forms are homotopy equivalent. So we just need to know that in each dimension there exists only finite number of connected compact Lie groups. This is indeed well known, see for example the references in the following mathoverflow answer: Classification of (compact) Lie groups