Is rejecting in polynomial time required for language to be in P?
Suppose you have a problem $A$, and a TM $M$ which accepts all the words $w \in A$ within a polynomial time bound $p(|w|)$, and diverge (or reject) otherwise.

Then, we can craft a new TM $N$, which on input $w$ performs the following:


simulate $M$ with input $w$ for $p(|w|)$ steps;
if $M$ accepts, we make $N$ accept as well
otherwise we make $N$ reject


It is immediate to see that $N$ runs in polynomial time: simulating a step of $M$ requires polynomial time on the $M$ configuration (whose size remains polynomial during the whole simulation), and we only need a polynomial number of steps.

Further, $N$ decides $A$ by construction.

TL;DR: if we know we can only accept a word within a time bound $t$, we can set an alarm on $t$. When the alarm rings, if we have not accepted before, we know it's a reject.

Conclusion: your alternative definition of $P$ is equivalent to the regular one. I would argue that it is harder to work with in proofs, though.