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Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".

But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.

Any suggestions or links?

For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:

For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:

1. 
   $G$ is reductive.


2. 
   $G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.


3. 
   $G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.

The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).

As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.

A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected normal unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.

But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.

A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.

Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.