The "assumption" in proof by induction

Your confusion is that you think you have proved something wrong!

You have proved that

"If there exists a largest positive integer then that largest positive integer is $1$"

That is a true statement.

Note that the statement $$ P \implies Q$$ is false only if $P$ is true and $Q$ is false

In your case $P$ is false , so your statement is true.

Indeed, a fallacious conclusion may be reached by valid deduction from an unjustifiable premise.

This is why the first step of induction is to prove that the predicate is justified for the base case; to ensure that we do not do that.

If $\mathcal P(0)$ is proven and for all natural numbers $n$ we can show that $\mathcal P(n)\to\mathcal P(n+1)$ is provable, then we may successively prove $\mathcal P(1)$, $\mathcal P(2)$, $\mathcal P(3)$, and so forth, by iterative applications of modus ponens.

$${\text{We may soundly prove }\mathcal P(1)\text{ from having proven }\mathcal P(0)\textit{ and }\mathcal P(0)\to\mathcal P(1)\\\text{We may soundly prove }\mathcal P(2)\text{ from having proven }\mathcal P(1)\textit{ and }\mathcal P(1)\to\mathcal P(2)\\\text{We may soundly prove }\mathcal P(3)\text{ from having proven }\mathcal P(2)\textit{ and }\mathcal P(2)\to\mathcal P(3)\\\text{And so forth, and so on,}\textit{ et cetera,}\ldots}$$

The assumption of the inductive hypothesis is valid because you have proven (in the first part of the proof by induction, the base case) that the statement $P$ holds for $n = n_0$. So you can think of it this way: initially, you only know that $P(n_0)$. But this, together with your proof of "if $P(n)$, then $P(n + 1)$", which implies "if $P(n_0)$, then $P(n_0 + 1)$", allows you to conclude that $P(n_0 + 1)$ by modus ponens. Another consequence of "if $P(n)$, then $P(n + 1)$ is "if $P(n_0 + 1)$, then $P(n_0 + 2)$", which then allows you to conclude that $P(n_0 + 2)$ by modus ponens. Continuing in this way, you can conclude $P(n_0 + k)$ for any natural number $k$ since any natural number $\ge n_0$ can be reached in this manner using a finite number of applications of modus ponens.