Trying to udnerstand textbook sample solution
You are correct; your "proof" is invalid; even if you show that $d$ is the correct value modulo $e$, that doesn't immediately imply that it is the correct value.  If you replace $d$ with $d+e$, that is also is correct modulo $e$, but is also obviously wrong (it is even).

Instead, here is a better line of reasoning:

Two integers $d, e$ work as the private, public exponents modulo a product of two primes $pq$ iff both the following hold:

$$de \equiv 1 \pmod{p-1}$$
$$de \equiv 1 \pmod{q-1}$$

What you need to show is:


That $d$ as specified in the above formula is actually an integer; that is, that $(p-1)(q-1)(e-1)+1$ is a multiple of $e$
That $de \equiv 1 \pmod{p-1}$
That $de \equiv 1 \pmod{q-1}$


None of these three should be difficult to prove.