We have
$$
S = \sum\limits_{\left\{ {\matrix{
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
{g_{\,1} + g_{\,2} + \, \ldots + \,g_{\,m} = W} \cr
} } \right.} {\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m} \cr} \right)
\left( \matrix{W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m} \cr} \right)}
$$
plus the added conditions
$$
\left\{ \matrix{
B + W = n \hfill \cr
g_{\,i} = x - k_{\,i} \hfill \cr
1 \le k_{\,i} \hfill \cr} \right.
$$
Leave apart for the moment the fact that $x=n/m, \; m | n$.
Because of the product, we can consider $0 \le ki$, and for the multinomial not to be null it shall be
$0 \le g_i$.
Thus the set of conditions becomes (if I understood them properly)
$$
\eqalign{
& \left\{ {\matrix{
{0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x} \cr
{g_{\,i} = x - k_{\,i} } \cr
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
{g_{\,1} + g_{\,2} + \, \ldots + \,g_{\,m} = W} \cr
} } \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ {\matrix{
{0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x} \cr
{g_{\,i} = x - k_{\,i} } \cr
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
{\left( {x - k_{\,1} } \right) + \left( {x - k_{\,2} } \right)
+ \, \ldots + \,\left( {x - k_{\,m} } \right) = W} \cr
} } \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ {\matrix{
{0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x} \cr
{g_{\,i} = x - k_{\,i} } \cr
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
{W + B = mx = n} \cr
} } \right. \cr}
$$
Under these conditions the summand can be rewritten as
$$
\eqalign{
& \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m} \cr} \right)
\left( \matrix{ W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m} \cr} \right) = \cr
& = \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{B!} \over {k_{\,1} !\,k_{\,2} !\, \cdots \,k_{\,m} !}}
{{\left( {mx - B} \right)!} \over {\left( {x - k_{\,1} } \right)!\,\left( {x - k_{\,2} } \right)!\, \cdots \,\left( {x - k_{\,m} } \right)!}} = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{x!} \over {k_{\,1} !\,\left( {x - k_{\,1} } \right)!}}
{{x!} \over {k_{\,2} !\,\left( {x - k_{\,2} } \right)!}} \cdots {{x!} \over {k_{\,m} !\,\left( {x - k_{\,m} } \right)!}} = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{x\left( {x - 1} \right)^{\,\underline {\,k_{\,i} - 1\,} } } \over {\left( {k_{\,i} - 1} \right)!\,}}} = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{\left( {x - 1} \right)^{\,\underline {\,k_{\,i} - 1\,} } } \over {\left( {k_{\,i} - 1} \right)!\,}}} =
{{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr k_{\,i} - 1 \cr} \right)} \cr}
$$
so that the sum becomes
$$
\eqalign{
& S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad = \cr
& = \sum\limits_{\left\{ {\matrix{
{0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x} \cr
{g_{\,i} = x - k_{\,i} } \cr
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
{g_{\,1} + g_{\,2} + \, \ldots + \,g_{\,m} = W} \cr
} } \right.}
{\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{B \cr k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m} \cr} \right)
\left( \matrix{ W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m} \cr} \right)} = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{
{\left( {1\, \le } \right)\,k_{\,i} \,\left( { \le \,x} \right)} \cr
{k_{\,1} + k_{\,2} + \, \ldots + \,k_{\,m} = B} \cr
} } \right.}
{\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr k_{\,i} - 1 \cr} \right)} } = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)} \cr
{j_{\,1} + j_{\,2} + \, \ldots + \,j_{\,m} = B - m} \cr
} } \right.}
{\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr j_{\,i} \cr} \right)} } \cr}
$$
Now, since
$$
\eqalign{
& \left( {\left( {1 + z} \right)^{x - 1} } \right)^{\,m}
= \prod\limits_{i = 1}^m {\left( {\sum\limits_{\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}
{\left( \matrix{ x - 1 \cr j_{\,i} \cr} \right)z^{\,j_{\,i} } } } \right)} = \cr
& = \sum\limits_{0\, \le \,s\, \le \,m\,\left( {x - 1} \right)}
{\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)} \cr
{j_{\,1} + j_{\,2} + \, \ldots + \,j_{\,m} = \,s} \cr
} } \right.}
{\left( {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr j_{\,i} \cr} \right)} } \right)z^{\,s} } } \cr}
$$
we conclude that
$$ \bbox[lightyellow] {
\eqalign{
& S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left[ {z^{\,B - m} } \right]\left( {1 + z} \right)^{m\left( {x - 1} \right)} = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left( \matrix{ m\left( {x - 1} \right) \cr B - m \cr} \right) = \cr
& = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {B - m} \right)!\left( {mx - B} \right)!}}\; = \cr
& = {{B!} \over {\left( {B - m} \right)!}}\;
{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}\;
= \left( \matrix{ B \cr m \cr} \right)\;
{{m!\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }} \cr}
}$$
--- in reply to your comment ---
Consider the following simple case
$$
\eqalign{
& \left( {1 + z} \right)^a \left( {1 + z} \right)^a
= \left( {\sum\limits_{k_1 } {\left( \matrix{ a \cr k_1 \cr} \right)z^{k_1 } } } \right)
\left( {\sum\limits_{k_2 } {\left( \matrix{ a \cr k_2 \cr} \right)z^{k_2 } } } \right) = \cr
& = \prod\limits_{i = 1}^2 {\left( {\sum\limits_{k_i }
{\left( \matrix{a \cr k_i \cr} \right)z^{k_i } } } \right)} = \cr
& = \sum\limits_{k_1 } {\sum\limits_{k_2 }
{\left( \matrix{ a \cr k_1 \cr} \right)\left( \matrix{ a \cr k_2 \cr} \right)z^{k_1 + k_2 } } }
= \sum\limits_s {\sum\limits_{k_1 + k_2 = s}
{\left( \matrix{a \cr k_1 \cr} \right)\left( \matrix{a \cr k_2 \cr} \right)z^s } } = \cr
& = \sum\limits_s {\left( {\sum\limits_{k_1 + k_2 = s} {\left( {\prod\limits_{i = 1}^2
{\left( \matrix{ a \cr k_i \cr} \right)} } \right)} } \right)z^s }
= \sum\limits_s {\left( \matrix{ 2a \cr s \cr} \right)z^s } \cr}
$$
which implies
$$
\sum\limits_{k_1 + k_2 = s} {\left( {\prod\limits_{i = 1}^2
{\left( \matrix{a \cr k_i \cr} \right)} } \right)}
= \left( \matrix{ 2a \cr s \cr} \right)
$$
and which is just the Vandermonde convolution.
Best Answer
When considering \begin{align*} \left(x_1+x_2+\cdots+x_n\right)^n \end{align*} we multiply $x_j$ with $t^j$ to track the index of the variable $x_j$. In the following we use the coefficient of operator $[t^k]$ to denote the coefficient of $t^k$ of a series $A(t)$ .
We show the following is valid for $n\geq 2$: \begin{align*} \color{blue}{[t^{2n-1}]\left(x_1t+x_2t^2+\cdots+x_nt^n\right)^n\Big|_{x_1=x_2=\cdots=x_n=1} =\binom{2n-2}{n-1}}\tag{1} \end{align*}
Comment:
In (2) we evaluate the expression at $x_1=x_2=\cdots=x_n=1$.
In (3) we factor out $t^n$ and apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$.
In (4) we apply the finite geometric series formula.
In (5) we skip all terms in the numerator $\left(1-t^n\right)^n=\sum_{j=0}^n\binom{n}{j}(-t^n)^j$ besides the constant term $1$ as the other terms do not contribute to $[t^{n-1}]$.
In (6) we make a binomial series expansion.
In (7) we select the coefficient of $t^{n-1}$.
In (8) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.