Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains, where $n\ge2$. Your conditions imply the following:
\begin{gather}p_{nn}=q_{nn},\\
p_{nj}<q_{nj}\text{ if }1\le j\le n-1.
\end{gather}
Hence, $1=\sum_{j=1}^n p_{nj}<\sum_{j=1}^n q_{nj}=1$, which is a contradiction (which makes any conclusion whatsoever true).
Added: In a comment, the OP suggested that the conjecture be modified by now assuming $n$ to be an absorbing state as well, so that the conditions become \begin{gather}p_{11}=q_{11}=p_{nn}=q_{nn}=1,\\ p_{ii}=q_{ii}\text{ if }1\le i\le n, \\ p_{ij}q_{ij}>0\text{ if }1<i<n,\\ p_{ij}>q_{ij}\text{ if }1<i<j\le n,\\ p_{ij}<q_{ij}\text{ if }1\le j<i<n. \end{gather}
The conjecture then becomes that \begin{gather}f_{P;ij}>f_{Q;ij}\text{ if }1<i<j\le n,\\ f_{P;ij}<f_{Q;ij}\text{ if }1\le j<i<n, \end{gather} where $f_{P;ij}$ is the probability that the first chain ever reaches $j$ from $i$, and $f_{Q;ij}$ is defined similarly.
This conjecture, too, is false in general. E.g., suppose that $n=5$, $$P=\frac1{20} \left( \begin{array}{ccccc} 20 & 0 & 0 & 0 & 0 \\ 4 & 4 & 4 & 4 & 4 \\ 2 & 2 & 2 & 12 & 2 \\ 4 & 4 & 4 & 4 & 4 \\ 0 & 0 & 0 & 0 & 20 \\ \end{array} \right),\quad Q=\frac1{20}\left( \begin{array}{ccccc} 20 & 0 & 0 & 0 & 0 \\ 7 & 4 & 3 & 3 & 3 \\ 12 & 4 & 2 & 1 & 1 \\ 5 & 5 & 5 & 4 & 1 \\ 0 & 0 & 0 & 0 & 20 \\ \end{array} \right). $$ Then $$f_{P;32}=\frac13\not<\frac{69}{283}=f_{Q;32}.$$