Here is a possible line of approach, lacking only the insight why
$(d+1)\frac{n}{b^d-1}$ must be $1$.
If the base $b$ representation of a number $n$ is cyclic of (exact) length $d\geq\lceil{\log_b n}\rceil$ (with inequality for leading zeros),
then the first $d$ consecutive multiples of $n$, $\{kn|1\leq k\leq d\}$,
exhaust (i.e. are in bijection with) all the cycles, which in turn
correspond bijectively with the repeating base-$b$ expansions of
$\frac{kn}{b^d-1}\in(0,1)$.
Their sum satisfies
$$
\frac{b^d-1}{b-1}s=
\frac{d(d+1)}{2}n
\qquad\text{where}\qquad
s=d\frac{b-1}{2}\frac{(d+1)n}{b^d-1}\in\mathbb{N}
$$
is the sum of the base-$b$ digits of $n$.
Since each digit is between $0$ and $b-1$,
$s$ must lie between $0$ and $d(b-1)$ inclusive.
However, for the sum $s$, the range must be exclusive (for $b>2$),
since otherwise the digits would all be the same ($0$ or $b-1$)
and $d$ would have to be $1$. Thus we have
$$0<\frac{s}{d}=\frac{b-1}{2}\frac{(d+1)n}{b^d-1}
Let's assume that $d>1$, and that $d$ is minimal, in the
sense that the $n$ is not a repeated or decomposable cycle:
$$
\nexists c\vert\;d,
\quad 1
We should also note that $s\equiv n\pmod{b-1}$,
i.e. that $\frac{n-s}{b-1}\in\mathbb{Z}$,
since each base-$b$ digit of $n$ remains fixed modulo $b-1$
when multiplied by its respective nonnegative power of $b$.
We actually want to show that
$$n=\frac{b^d-1}{d+1}
\qquad\text{and that}\qquad
t=\frac{b^d-1}{n}=d+1\in\mathbb{N}$$
is in fact prime with $b$ as primitive root.
Perhaps there is a good argument why $s=d\frac{b-1}{2}$ lies exactly in the middle (the expected value) of the prescribed range or, equivalently,
that the average value of the digits of $n$ base $b$ is $\frac{b-1}{2}$, or
that the first noncyclic multiple of $n$ (which we know is the $d+1^\text{th}$) satisfies $(d+1)n=b^d-1$ (which is $b-1$ times the repunit of length $d$).
Certainly we know that the sequence $\{kn\}_{k=1}^d$ is increasing and bounded by $b^d$ (since each term has at most $d$ digits base $b$). Therefore, they must correspond with the lexicographic ordering of cyclically shifted length-$d$ strings starting with $n$, zero-padded if necessary (i.e. if $b^{d-1}>n$). And since $(d+1)n=dn+n$ is the sum of the smallest and largest of these cyclic shifts, its leading digit must also be the sum of the smallest and largest digits of $n$ (unless a carry increases the sum to $\geq b^d$).
If we need to resort to an examination in more detail of the products $\{kn\}_{k=1}^d$ and their relation to base-$b$ shifts of $n$, we need not resort to naming $n$'s digits. We can in stead rely on the division algorithm, and note that if
$$
n=q_k b^k+r_k
\quad\text{with}\quad
\left\{\begin{matrix}
q_k=\lfloor{b^{-k}n}\rfloor,\\
r_k=n-b^kq_k
\end{matrix}\right.
\quad\text{for}\quad
0\leq k\leq d
\quad\text{and if}\quad
n_k=r_k b^{d-k}+q_k,
$$
then $\{n_k\}_{k=1}^{d-1}$
is a permutation of $\{nk\}_{k=1}^{d-1}$.
Note that if $n$ and $n_k$ are identified with
strings of $d$ letters in the alphabet $\{0,\dots,b-1\}$,
then $n_k=\text{right}(n,k)+\text{left}(n,d-k)$,
where the plus symbol here indicates string concatenation and the
right and left functions are familiar from some programming languages,
since $q_k$ and $r_k$ are the left $d-k$ and
right $k$ digits of $n$ base $b$ respectively.
Once we can establish that $n=\frac{b^d-1}{d+1}$,
we would have that $b^d\equiv 1\pmod t$.
From here we could argue that $b$ must have order $d$ modulo $t$
using the minimality of $d$: if $1
But this would prove the result, since then
$t-1=d=\text{ord}_t(b)\leq\phi(t)\leq t-1$,
i.e. we would have sandwiched Euler's totient function
$\phi(t)$ into attaining its theoretical maximum,
which only occurs when $t$ is prime, while on the other hand,
the order of any element $b$ mod $t$ only attains $\phi(t)$
when the element $b$ is a primitive root,
i.e. a generator of $(\mathbb{Z}/t\mathbb{Z})^*$.
Finally (just for fun), note that a start at factoring $n$ for $d>1$ is
$$
n=\frac{b^d-1}{t}=\frac{1}{t}\prod_{0cyclotomic polynomial of degree $c$,
and the product above will be a partial factorization with $\tau(d)$ terms,
one of which must be divisible by the prime $t$,
where $\tau(d)$ is the number of positive divisors of $d$.