First of all, re Stars and Bars Theory, the two methods that you used do give the same answer.
That is $p_{10}(100)$ bijects to determining the number of non-negative integer solutions to $x_1 + \cdots + x_{10} = 90.$ So, both methods do yield
$$\frac{\binom{99}{9}}{\binom{109}{9}}. \tag1 $$
Despite that, the answer in (1) above is still wrong. The point is that, as you indicated, each individual solution to (for example) $x_1 + \cdots + x_{10} = 100$ is not equally likely to occur.
Edit
Actually, my analysis of the two approaches taken by the OP (i.e. Original Poster) was careless. In his 2nd approach, he specifically assumed that the boxes were to be regarded as indistinguishable. So, my analogy of (for example) enumerating the number of positive integer solutions to $x_1 + \cdots + x_{10} = 100$ is not valid.
Just to emphasize the point that even regarding the boxes as indistinguishable is still the wrong way to go, consider the following simplification.
Suppose that you have $6$ balls to be distributed to $3$ blns, and let $(b_1, b_2, b_3)$ denote a solution where bin-1 receives $b_1$ balls, bin-2 receives $b_2$ balls, and bin-3 receives $b_3$ balls.
Consider the various solutions to $b_1 + b_2 + b_3 = 6.$
First, assume that the bins are distinguishable. Then, there are $~\displaystyle \binom{6}{2}\binom{4}{2} = 90~$ distinct ball distributions that are represented by the $(2,2,2)$ solution and $~\displaystyle
\binom{6}{3}\binom{3}{2} = 60~$ distinct ball distributions that are represented by the $(3,2,1)$ solution.
So, the $(2,2,2)$ and $(3,2,1)$ solutions do not represent an equal number of ball distributions into the bins.
Now, consider what happens if an attempt is made to construe the bins as indistinguishable. It is as if the bins were first labeled, the balls were distributed to the bins, and then the labels were removed from the bins.
With the bins being considered indistinguishable, the $(2,2,2)$ solution only represents $(b_1,b_2,b_3) = (2,2,2)$. In contrast, the $(3,2,1)$ solution actually represents $(b_1, b_2, b_3)$ being any element in
$$\{(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)\}.$$
So, by construing the bins as indistinguishable, and then enumerating the different partitioning possibilities, you have that
the two partitioning solutions of $(2,2,2)$ and $(1,2,3)$ actually represent distributions sets whose sizes are in the proportions of $[90::360]$.
This means that if you are going to enumerate the partitions, where the bins are considered distinguishable, then the $(2,2,2)::(1,2,3)$ solutions must be given relative weights of $[90::60]$.
If instead, you enumerate the partitions where the bins are considered indistinguishable, then the $(2,2,2)::(1,2,3)$ solutions must be given relative weights of $[90::360]$.
Another approach that is no good is
$$1 - \frac{\binom{10}{1}\times (9)^{(100)}}{(10)^{(100)}}. \tag2 $$
The flawed idea re (2) above is that in the numerator, first you select a bin to be left empty. Then, each of the $(100)$ balls has $9$ choices for the other bins. The reason that (2) above is no good is because of over-counting. That is, if you leave bin-1 empty, included in the $(9)^{(100)}$ computation will be occurrences where bin-2 was left empty. So, the times when bin-1 and bin-2 are both left empty is counted twice.
It is unclear to me whether there is an elegant approach that involves either direct computation, recursion, or generating functions. My go-to strategy, in this situation, is Inclusion-Exclusion.
See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
I am going to adopt the syntax of the 2nd link above.
$T_0$ will equal $(10)^{(100)}$, which will also serve as the denominator, just as it did in (2) above.
Let $S$ denote the set of all possible distributions of the balls into the $10$ bins. Then, $|S| = T_0$.
For $k \in \{1,2,\cdots,k\}$, let $S_k$ denote the subset of $S$, where bin-k was left empty.
Then, the desired computation will be
$$\frac{|S| - |S_1 \cup \cdots \cup S_k|}{|S|}. \tag3 $$
Let $T_1$ denote $|S_1| + \cdots + |S_{10}|.$
For $r \in \{2,3,\cdots,10\}$,
let $T_r$ denote $~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 10} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|.$
That is, $T_r$ represents the sum of $~\displaystyle \binom{10}{r}$ terms.
Then, in accordance with Inclusion-Exclusion Theory, the desired computation, which is represented in (3) above, will equal
$$\frac{\sum_{r=0}^{10} (-1)^r T_r}{T_0}, \tag4 $$
where $~\displaystyle T_0 = (10)^{(100)}.$
So, in order to resolve (4) above, all that remains is to evaluate $T_1, T_2, \cdots, T_{10}.$
$\underline{\text{Computation of} ~T_1:}$
$\displaystyle |S_1| = 9^{(100)}.$
By reasons of symmetry, $|S_1| = |S_2| = \cdots = |S_{10}|.$
Therefore,
$$T_1 = \binom{10}{1} 9^{(100)}. \tag5 $$
$\underline{\text{Computation of} ~T_2:}$
$\displaystyle |S_1 \cap S_2| = 8^{(100)}.$
By reasons of symmetry, for any $i,j \in \{1,2,\cdots,10\} ~: i \neq j$,
you have that $|S_i \cap S_j| = |S_1 \cap S_2|.$
Therefore,
$$T_2 = \binom{10}{2} (10-2)^{(100)}. \tag6 $$
$\underline{\text{Computation of} ~T_r: 3 \leq r \leq 9}$
$\displaystyle |S_1 \cap S_2 \cap \cdots \cap S_r| = (10-r)^{(100)}.$
By reasons of symmetry, there will be $~\displaystyle \binom{10}{r}~$ identical terms.
Therefore,
$$T_r = \binom{10}{r} (10-r)^{(100)}. \tag7 $$
$\underline{\text{Final Computation}}$
Clearly, $T_{10} = 0$, because it is impossible to distribute the $(100)$ balls into the bins, with the result that all of the bins are empty.
Therefore, the final computation is
$$\frac{(10)^{(100)} + \sum_{r=1}^{9} \left[(-1)^r \times \binom{10}{r} \times (10-r)^{(100)}\right]}{(10)^{(100)}}.$$
