Here's the full combinatorial argument. Raney's lemma was about half of it, I'd say. The argument shows that $S(n,2) = \binom{3n-1}{n} \frac{3}{3n-1}$ (which is equivalent to the two formulations I give in the question). At the end I'll discuss the generalization to the $r$ case.
Intro.
Consider the sequences with $2n$ occurrences of $-1$ (i.e., $2n$ heads) and $n$ occurrences of $+2$ (i.e., $n$ tails). We want to show that the number of these sequences with all partial sums nonzero is $\binom{3n-1}{n} \frac{3}{3n-1}$. The complete sum and empty sum are clearly $0$, so "partial sum" excludes those two cases. The sequences we want to count can be split into three groups: (1) all partial sums positive, (2) all partial sums negative, (3) some partial sums positive and some negative.
Group 1: The number of these sequences with all partial sums positive is $\binom{3n-1}{n} \frac{1}{3n-1}$.
This is the part that uses Raney's lemma. If all partial sums are positive, the last element in the sequence must be $-1$. Thus we want to count the number of sequences with $2n-1$ occurrences of $-1$ and $n$ occurrences of $+2$ that add to $+1$ and have all partial sums positive. Ignoring the partial sums restriction, there are $\binom{3n-1}{n}$ such sequences. If we partition these $\binom{3n-1}{n}$ sequences into equivalence classes based on cyclic shifts, Raney's lemma says that exactly one sequence in each equivalence class has all partial sums positive. Because there are $3n-1$ elements in each sequence there $3n-1$ sequences with the same set of cyclic shifts, and so there are $3n-1$ sequences in each equivalence class. Thus the number of sequences in Group 1 is $\binom{3n-1}{n} \frac{1}{3n-1}$.
Group 2: The number of these sequences with all partial sums negative is also $\binom{3n-1}{n} \frac{1}{3n-1}$.
To see this, just reverse the sequences counted in Part 1.
Group 3: The number of these sequences with some positive partial sums and some negative partial sums is, yet again, $\binom{3n-1}{n} \frac{1}{3n-1}$.
This one is a little trickier. First, because of the $-1$'s, it is not possible to switch from positive partial sums to negative partial sums. Thus any sequence counted here must have exactly one switch: from negative partial sums to positive partial sums. The switch must occur at some point where the partial sum is $-1$ and the next element is $+2$. Thus we have some sequence $(a_1, \ldots, a_m, +2, a_{m+2}, \ldots, a_n)$ where the sums $a_1, a_1 + a_2, \ldots, a_1 + \ldots + a_m$ are all negative. Consider the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$. Since $+2 + a_m + \ldots + a_1 = 1$, and $a_k + a_{k-1} + \ldots + a_1 < 0$ for all $k$, $1 \leq k \leq m$, it must be the case that $+2 + a_m + \cdots + a_{k+1} > 1$ for all $k$, $1 \leq k \leq m-1$. So the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$ is in Group 1. To see that this mapping is a bijection, note that any sequence in Group 1 must start with $+2$ and have a first time that a partial sum is equal to $+1$. Thus this transformation is reversible.
Summing up.
Putting Groups 1, 2, and 3 together we see that the total number of sequences we want to count is $\binom{3n-1}{n} \frac{3}{3n-1}$.
Generalization to the $r$ case.
The arguments for Group 1 and Group 2 adapt in a straightforward manner for general $r$; we get $\binom{(r+1)n-1}{n} \frac{1}{(r+1)n-1}$ for both. Sequences in Group 3 can still only switch once - from negative partial sums to positive partial sums. But Group 3 can be broken up into subgroups depending on whether the first partial sum to become positive is $+1, +2, \ldots, r-1$. Using transformations like the one described above for Group 3 in the $r = 2$ case, we can show that there is a bijection between each of these $r-1$ subgroups and Group 1. Thus there are $\binom{(r+1)n-1}{n} \frac{r-1}{(r+1)n-1}$ sequences in Group 3. In total, then, $S(n,r) = \binom{(r+1)n-1}{n} \frac{r+1}{(r+1)n-1}$.
Final comment.
All of this is may be easier to visualize in terms of paths from $(0,0)$ to $((r+1)n,(r+1)n)$ that use right steps of size $1$ and up steps of size $r$ and that do not step on the diagonal except at $(0,0)$ and $((r+1)n,(r+1)n)$. (They can cross the diagonal, though.) I found it easier to discuss the partial sums interpretation, though, given the difficulty of creating such graphics on this forum.
We seek to show with Bernoulli numbers, Catalan numbers, Eulerian
numbers and the Factorial counterpart of Leibniz's harmonic triangle:
$$\frac{B_n}{C_n} = \sum_{k=0}^n (-1)^k
\frac{k! (n-k)!}{(n+1)^\overline{n}}
\left\langle {n \atop k} \right\rangle.$$
Seeing that $(n+1)^\overline{n} = (2n)^\underline{n}$ we get the
equivalent statement
$$B_n = \frac{1}{(n+1)!}
\sum_{k=0}^n (-1)^k k! (n-k)!
\left\langle {n \atop k} \right\rangle
\\ =
\sum_{k=0}^{n} (-1)^k \mathrm{B}(k+1, n-k+1)
\left\langle {n \atop k} \right\rangle
\\ = \int_0^1 u^n
\sum_{k=0}^{n} (-1)^k \frac{(1-u)^k}{u^k}
\left\langle {n \atop k} \right\rangle \; du.$$
We suppose as given that
$$\sum_{k=0}^n \left\langle {n \atop k} \right\rangle t^k
= n! [x^n] \frac{t-1}{t-\exp((t-1)x)}.$$
This yields
$$\int_0^1 u^n n! [x^n]
\frac{(u-1)/u-1}{(u-1)/u-\exp(((u-1)/u-1)x)} \; du
\\ = n! [x^n] \int_0^1
\frac{(u-1)-u}{(u-1)-u\exp((u-1-u)x)} \; du
\\ = n! [x^n] \int_0^1
\frac{1}{1-u+u\exp(-x)} \; du
\\ = n! [x^n] \left.
\frac{\log(1+(\exp(-x)-1)u)}{\exp(-x)-1}
\right|_0^1
\\ = n! [x^n] \frac{(-x)}{\exp(-x)-1}
= n! [x^n] \frac{x \exp(x)}{\exp(x)-1}.$$
Now the EGF of the Bernoulli numbers $\{B^{-{}}_n\}$ is
$$\frac{x}{\exp(x)-1}$$
so that we get for the revised EGF of $\{B^{+{}}_n\}$ by Luschny, Knuth
et al. correcting the value at $n=1$ from $-\frac{1}{2}x$ to $\frac{1}{2}
x$
$$\frac{x}{\exp(x)-1}+x
= \frac{x+x(\exp(x)-1)}{\exp(x)-1}
= \frac{x\exp(x)}{\exp(x)-1}$$
as claimed.
What we have here is another match score in favor of $\{B^{+{}}_n\}$
as defined in Luschny's Bernoulli Manifesto.
Best Answer
Noticing that $\binom{a}{b}=\frac{(a)_b}{b!}, $ you can write your expression as $$\binom{x+n-1}{n}=\sum _{k=1}^n\binom{n-1}{n-k}\binom{x}{k}$$ which is Vandermonde, so the standard combinatorial proof of Vandermonde suffices.
Notice that the explicit formula comes from shuffling the elements of $[n]$ in a line, using stars and bars to see that $\binom{n-1}{n-k}$ is the number of ways to partition the $n$ into positive $k$ parts $a_1+a_2+\cdots +a_k=n$ where order matters, and so you partition your line using first $a_1$ numbers, then $a_2$ until you get the $n$ numbers, and taking out the order of the $k$ parts. For example: Given $a_1=4,a_2=2,a_3=1,a_4=4$ then $$\underbrace{1\,9\,10\,2}_{a_1}\,\underbrace{5\,6}_{a_2}\,\underbrace{4}_{a_3}\,\underbrace{7\,3\,8}_{a_4}\text{ gives the ordered partition.}$$
It suffices to show it just for $x$ an integer because this are polynomials on $x$ and if you have two polynomials $P_1(x)=P_2(x)$ for $x\in \mathbb{N}$ then $$P_1(x)-P_2(x)=0$$ implies that the polynomial on the LHS has infinite roots, that is just possible if the polynomial on the left is strictly $0.$