We have 3 assumptions:
1) We have two different ordering relations $ \leq_a $ and $ \leq_b $.
2) Given arbitrary sets $S_1, S_2 $, we know $ S_1 \leq_a S_2 \rightarrow S_1 \leq_b S_2 $.
3) We also know $\leq_a$ is a preorder on $\mathbb{N} $.
?) Does this imply that $\leq_b$ is a preorder on $\mathbb{N} $?
No it does not, let $\leq_b$ be the relation on $\mathbb N$ so that $x\leq_b y$ always except if $x=1$ and $y=0$.
Let $\leq_a$ be the usual order on $\mathbb N$. Then clearly $\leq_a$ is a preorder and clearly if $x\leq_ay$ then $x\leq_by$. But $\leq_b$ is not a preorder since $1\leq_b 2$ and $2\leq_b0$ but $1\not \leq_b 0$.