Assume that there are positive solutions. Then there is a solution $(a,b,c,d)$ with the smallest possible value $abcd > 0$.
Considering the equation mod $2$, you see that $d$ is even, so $d = 2\delta$, and division by $2$ gives $4a^4 + 2b^4 + c^4 = 8\delta^4$. Now mod $2$ we find that $c$ is even, so $c = 2\gamma$. Going on like this, also $b = 2\beta$ and $a = 2\alpha$ are even, and we get the solution $(a/2,b/2,c/2,d/2)$ which contradicts the minimality condition on $(a,b,c,d)$.
The axioms you listed (P1-P5') are not equivalent to Peano's. Replacing the (weak) induction axiom with the well-ordering axiom gives a weaker theory. The well-ordered sets that are not order-isomorphic to the natural numbers still obey the well-ordering axiom.
Before I come back to the trichotomy question, let's recall the role of induction in Peano's axioms. Like all the other axioms, induction is chosen so that the resulting theory may describe as well as possible the natural numbers. Induction, specifically, is there to exclude certain undesirable models.
For instance, without induction, one could easily build a "non-standard" model of arithmetic that, besides the natural numbers, contained two more elements, call them $\alpha$ and $\beta$, such that $s(\alpha) = \beta$ and $s(\beta) = \alpha$. Such a model would satisfy P1-P4, but induction rules it out.
If Peano's axioms included strong induction instead of weak induction, the resulting theory would allow models whose order type is not $\omega$. For instance, $W = \{0,1\} \times \mathbb{N}$, with lexicographic ordering, satisfies P1-P5'. (In particular, it is a well-order.)
On the other hand, weak induction does not "work" on $W$, because starting from $(0,0)$, which is the least element of $W$, and working up to $(0,1), (0,2)$ and so on, one never reaches $(1,0)$. One could "prove" by weak induction that all elements of $W$ have first component equal to $0$. For $W$ one needs transfinite induction, which is essentially strong induction.
Now, for the trichotomy property. Note that Mendelson introduces $x < y$ as a "purely abbreviational definition" in terms of $+$. Therefore, $<$ must be proved a total order. For that, induction is used; specifically, to show that the trichotomy property holds.
When proving that a well-ordered set satisfies the strong induction principle, the ordering of the set is supposed to be given, and to be a strict total order. No property of strict total orders needs to be proved.
Best Answer
Hint $\ $ Suppose for contradiction that some natural $> 1$ hs no prime factorization, and let $n$ be minimal such. Then $ n$ is not prime, so it has a proper factorization $\ n = ab.\ $ Since $a,b < n,\, $ what can you deduce about $\,a,b\,$ from the minimality hypothesis on $\,n\,$? $ $ Finally, what does the prior deduction imply about $\,n,\,$ given that is the product of $\,a\,$ and $\,b\,$?