Here’s a possible algorithm.
Construct a graph $G’$ with $|V|^2$ vertices where each vertex is labeled with the pair $(a, b)$ with $a, b$ being vertices in $G$. Then, construct all possible edges $(a, b) to (c, d)$ in $G’$ where two edges $c to a$ (note the order) and $b to d$ exist in $G$ with the same color.
Then to find an odd-length palindrome you simply check if $(s, t)$ is reachable from any $(x, x)$ in $G’$. To find an even-length palindrome check if $(s, t)$ is reachable from any $(a, b)$ in $G’$ under the condition that $a to b$ exists (regardless of color) in $G$.
The idea is to extend the palindrome from the middle out. The vertices $(l, r)$ in $G’$ represent the leftmost and rightmost symbol in our palindrome, and the way we constructed the edges in $G’$ corresponds to all valid ways to extend the palindrome.