TL;DR: The answer to the title question is, "No, not always."
But when the equation is written as stated in the question,
it is possible to conclude that $x = yz/n$ is a solution
without needing to rewrite the equation in a "simplified" form.
This is simple to show even in the case where $a_k = b_k = 0$ in one or more terms.
Further details:
You don't need to divide by zero.
You can simply require that $a_k = \dfrac zn b_k$
for $k \in \{1, \ldots, n\}$, which is still defined on both sides and true when $a_k$ and $b_k$ are both zero.
What will be a problem is if $b_k = -y$ for some $b_k$, because then you actually do have a division by zero. Of course, when the problem is given with explicit numbers (like the problem from the other question), the division by zero will be quite obvious.
Note that
\begin{multline}
\frac{x+a_1}{y+b_1} + \frac{x+a_2}{y+b_2} + \cdots + \frac{x+a_n}{y+b_n} = \\
\left(\frac{1}{y+b_1} + \frac{1}{y+b_2} + \cdots + \frac{1}{y+b_n}\right) x
+ \left(\frac{a_1}{y+b_1} + \frac{a_2}{y+b_2} + \cdots + \frac{a_n}{y+b_n}\right),
\end{multline}
so when we set this quantity equal to $z$ we have an equation of the form
$mx + b = z$, which has exactly one solution if $m\neq 0$ and has either no solution or infinitely many solutions if $m = 0$.
If $m = 0$ we cannot conclude that $\dfrac xy = \dfrac zn$.
For example, consider
$$
\frac{x+1}{1+1} + \frac{x-3}{1-3} = 2.
$$
This satisfies the condition that $a_k = \dfrac zn b_k$
for $k \in \{1, \ldots, n\}$,
and every real $x$ is a solution.
If $y + b_k$ has the same sign for every $k$ then $m \neq 0$, but that is not a necessary condition. If $m = 0$ then it is sufficient to make any change in exactly one of the $a_k$ values in order to ensure that $m \neq 0$.
If you want a formal proof, I think your best bet is first to prove that each term of the sum is equal to $\frac zn$ when $x = \frac zn y$, essentially as you did already.
Then add up the $n$ equal terms.
But it must be all $n$ terms, not just $n - 1$ terms.
Try something like this:
If $a_k = \frac zn b_k$ and $x = \frac zn y$,
then
$$ x + a_k = \frac zn y + \frac zn b_k = \frac zn (y + b_k). $$
Provided that $y + b_k \neq 0$, we can divide each side by $y + b_k$
to show that
$$
\frac{x + a_k}{y + b_k} = \frac zn.
$$
I think now I really must emphasize that
$$
n\left(\frac xy\right) = z
$$
is not a generally valid simplification of the equation
$$
\frac{x+a_1}{y+b_1} + \frac{x+a_2}{y+b_2} + \cdots \frac{x+a_n}{y+b_n} = z.
$$
These two equations are equivalent (have the same solutions for $x$)
only under the additional condition that
$$
\frac{1}{y+b_1} + \frac{1}{y+b_2} + \cdots + \frac{1}{y+b_n} \neq 0. \tag{C}
$$
There is no valid "simplification" from one equation to the other unless you include this condition as a requirement for the simplification.
To solve the problem in general, it is sufficient to recognize that
$x = \frac zn y$ is a solution and then to check whether there are any other solutions. For the check for other solutions, you have two options:
a) Verify Inequality (C). This can be easy to do; for example, if all denominators are positive, you have a unique solution.
b) Check just one other value of $x$. You can check $x = 0$ or $x = -a_1$ or any value of $x$ you find convenient. If the equation is still true, it's true for all $x$; otherwise, $x = \frac zn y$ is the unique solution.
There are, of course, innumerable other ways of invalidating the solution method by making minor adjustments to the problem, such as combining multiple terms into one or multiplying $x$ by some coefficient other than $1$ in one of the terms. But every one of those creates an equation that is not in the form
$$
\frac{x+a_1}{y+b_1} + \frac{x+a_2}{y+b_2} + \cdots + \frac{x+a_n}{y+b_n} = z
\tag{P}
$$
where there are exactly $n$ terms on the left-hand side.
So we can lump all of these exceptions into one general exception:
"if the equation is not in the exact form of Equation (P)."
There are also many ways to modify the solution so that some of these exceptions are no longer exceptions. For example,
let $C = \sum_1^n c_k$ where each $c_k$ is a known fixed value and $C \neq 0$,
let $a_k = \dfrac{c_k z}{C} b_k$ for $k \in \{1, \ldots, n\}$
where $a_k$ and $b_k$ are known fixed values, let $y$ and $z$ be known fixed values, let $y + b_k \neq 0$ for $k \in \{1, \ldots, n\}$, and let
$$
\frac{c_1 x + a_1}{y + b_1} + \frac{c_2 x + a_2}{y + b_2} + \cdots
+ \frac{c_n x + a_n}{y + b_n} = z. \tag{Q}
$$
Then if $x = \dfrac{z}{C} y$ for $k \in \{1, \ldots, n\}$,
we find that for each $k$,
$$
c_k x + a_k = c_k \frac zC y + \frac{c_k z}C b_k = c_k \frac zC (y + b_k)
$$
and dividing by $y + b_k$ on each side, we find that
$$
\frac{c_k x + a_k}{y + b_k} = c_k \frac zC.
$$
Then the left-hand side of Equation (Q) is
$$
\sum_1^n \frac{c_k x + a_k}{y + b_k}
= \sum_1^n c_k \frac zC = \frac zC \sum_1^n c_k = \frac zC C = z,
$$
and therefore $x = \dfrac{z}{C} y$ is a solution of Equation (Q).
Equation (P) is just Equation (Q) with $c_k = 1$ for every $k$.
Now further suppose that we replace the condition $y + b_k \neq 0$ with the condition $d_k y + b_k \neq 0$ for $k \in \{1, \ldots, n\}$,
where each $d_k$ is a known fixed quantity. Let
$$
\frac{c_1 d_1 x + a_1}{d_1 y + b_1} + \frac{c_2 d_2 x + a_2}{d_2 y + b_2}
+ \cdots + \frac{c_n d_n x + a_n}{d_n y + b_n} = z. \tag{R}
$$
Through methods similar to the ones before, we can show that
if $x = \dfrac{z}{C} y$ then
$$
\frac{c_k d_k x + a_k}{d_k y + b_k} = c_k \frac zC
$$
and therefore $x = \dfrac{z}{C} y$ is a solution of Equation (R).
By this point, however, it may be getting difficult to recognize whether an
equation actually is in the form (R) when each of the quantities
$c_k d_k$ and $d_k y + b_k$ has been written as a single explicit number
such as $30$ or $2025$.
At some point, as we generalize the method, I think we will reach a point where the generalization is no longer useful or interesting.
