Here are a few examples from graph-Ramsey theory. In the first pair of examples, the Ramsey version and density version are essentially as far apart as one can get. In the last two pairs of examples, the two versions coincide. Now I wonder if there is an example from graph-Ramsey theory where the bound from the density version is strictly stronger than the Ramsey version? But in general, it seems to me that your question could be narrowed down by simply asking for results in which the density version implies the Ramsey version, since those seem to be more rare.
1R) In every 2-coloring of $K_n$, there is a monochromatic connected subgraph on $n$ vertices. (Folklore)
1D) Every graph on $n$ vertices with at least $\binom{n-1}{2}+1$ edges is connected. (Folklore)
2R) In every 2-coloring of $K_n$, there is a monochromatic path on at least $2n/3$ vertices. (Gerencsér, Gyárfás)
2D) Every graph one $n$ vertices with at least $\frac{2}{3}\binom{n}{2}$ edges has a path on at least $2n/3$ vertices. (Erdős, Gallai)
3R) In every 2-coloring of $K_n$, there is a monochromatic matching covering at least $2n/3$ vertices. (Cockayne, Lorimer)
3D) Every graph on $n$ vertices with at least $\frac{5}{9}\binom{n}{2}$ edges has a matching covering at least $2n/3$ vertices. (Erdős, Gallai)
4R) In every 2-coloring of $K_n$, there is a monochromatic copy of every tree $T$ with at most $n/2+1$ vertices. Furthermore, there are trees with more than $n/2+1$ vertices for which this is not true. (Burr-Erdős conjecture, solved for large $n$ by Zhao)
4D) Every graph on $n$ vertices with at least $\frac{1}{2}\binom{n}{2}$ edges contains every tree with at most $n/2+1$ vertices. (Erdős, Sós conjecture)
5R) In every $r$-coloring of $K_{n,n}$, there is a monochromatic connected subgraph on at least $2n/r$ vertices and this is essentially best possible. (Gyárfás)
5D) Every balanced bipartite graph on $2n$ vertices with at least $n^2/r$ edges has a connected subgraph on at least $2n/r$ vertices and this is also best possible. (Gyárfás; Mubayi; Liu, Morris, Prince)