Here is a two-step approach. At first we transform the identity into an equivalent one which contains a finite sum only. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}
Since
\begin{align*}
\binom{T}{B}=\binom{T}{T-B}=\frac{T}{T-B}\binom{T-1}{T-B-1}=\frac{T}{T-B}\binom{T-1}{B}
\end{align*}
we want to show
\begin{align*}
\color{blue}{T\binom{T-1}{B}=\sum_{l=1}^{\infty}\binom{T+l-1}{B-1}\left(1-\frac{B}{T}\right)^ll}\tag{2}
\end{align*}
Putting $\alpha:=1-\frac{B}{T}$ the right-hand side of (2) can be written as>
\begin{align*}
\color{blue}{\sum_{l=1}^{\infty}}&\color{blue}{\binom{T+l-1}{B-1}\alpha^ll}\\
&=\sum_{l=1}^{\infty}[z^{B-1}](1+z)^{T+l-1}\alpha^ll\tag{4.1}\\
&=[z^{B-1}](1+z)^{T-1}\sum_{l=1}^{\infty}l\alpha^l(1+z)^l\\
&=[z^{B-1}](1+z)^{T}\frac{d}{dz}\sum_{l=1}^{\infty}\alpha^l(1+z)^l\\
&=[z^{B-1}](1+z)^T\frac{d}{dz}\left(\frac{1}{1-\alpha(1+z)}-1\right)\\
&=[z^{B-1}]\frac{\alpha(1+z)^T}{\left(1-\alpha(1+z)\right)^2}\\
&=[z^{B-1}]\frac{\alpha(1+z)^T}{(1-\alpha)^2\left(1-\frac{\alpha z}{1-\alpha}\right)^2}\\
&=\frac{\alpha}{(1-\alpha)^2}[z^{B-1}](1+z)^T
\sum_{j=0}^{\infty}\binom{-2}{j}\left(-\frac{\alpha z}{1-\alpha}\right)^j\tag{4.2}\\
&=\frac{\alpha}{(1-\alpha)^2}[z^{B-1}]
\sum_{i=0}^T\binom{T}{i}z^i\sum_{j=0}^{\infty}(j+1)\left(\frac{\alpha z}{1-\alpha}\right)^j\tag{4.3}\\
&=\frac{\alpha}{(1-\alpha)^2}[z^{B-1}]
\sum_{n=0}^{\infty}\sum_{j=0}^n(j+1)\left(\frac{\alpha}{1-\alpha}\right)^j\binom{T}{n-j}z^n\tag{4.4}\\
&=\frac{\alpha}{(1-\alpha)^2}\sum_{j=0}^{B-1}\binom{T}{B-1-j}(j+1)\left(\frac{\alpha}{1-\alpha}\right)^j\tag{4.5}\\
&=\frac{1}{1-\alpha}\sum_{j=0}^{B-1}\binom{T}{B-(j+1)}(j+1)\left(\frac{\alpha}{1-\alpha}\right)^{j+1}\\
&\,\,\color{blue}{=\frac{1}{1-\alpha}\sum_{j=1}^{B}\binom{T}{B-j}j\left(\frac{\alpha}{1-\alpha}\right)^j}\tag{4.6}\\
\end{align*}
Comment:
In (4.1) we use the coefficient of operator according to (1).
In (4.2) we use the binomial series expansion.
In (4.3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (4.4) we multiply out the series.
In (4.5) we select the coefficient of $z^{B-1}$.
In (4.6) we shift the index $j$ by 1 and start with $j=1$.
Since $\frac{\alpha}{1-\alpha}=\frac{T}{B}\left(1-\frac{B}{T}\right)=\frac{T}{B}-1$, we obtain from (4.6) the following finite sum identity equivalent with (2):
\begin{align*}
\color{blue}{B\binom{T-1}{B}=\sum_{j=0}^B\binom{T}{B-j}j\left(\frac{T}{B}-1\right)^j}\tag{5}
\end{align*}
Now in a second step we can considerably simplify the right-hand sum of (5). Putting $z:=\frac{T}{B}-1$ we obtain
\begin{align*}
\color{blue}{\sum_{j=0}^B}&\color{blue}{\binom{T}{B-j}jz^j}\\
&=\sum_{j=0}^B\binom{T}{j}(B-j)z^{B-j}\tag{6.1}\\
&=B\sum_{j=0}^B\binom{T}{j}z^{B_j}-\sum_{j=1}^B\binom{T}{j}jz^{B-j}\\
&=B\sum_{j=0}^B\binom{T}{j}z^{B-j}-T\sum_{j=1}^B\binom{T-1}{j-1}z^{B-j}\tag{6.2}\\
&=B\left(\sum_{j=0}^B\binom{T-1}{j}z^{B-j}+\sum_{j=1}^B\binom{T-1}{j-1}z^{B-j}\right)\\
&\qquad\qquad-\frac{T}{z}\sum_{j=0}^{B-1}\binom{T-1}{j}z^{B-j}\tag{6.3}\\
&=B\left(\sum_{j=0}^B\binom{T-1}{j}z^{B-j}+\frac{1}{z}\sum_{j=0}^{B-1}\binom{T-1}{j}z^{B-j}\right)\\
&\qquad\qquad-\frac{T}{z}\sum_{j=0}^{B-1}\binom{T-1}{j}z^{B-j}\tag{6.4}\\
&=B\binom{T-1}{B}+\left(B-(T-B)\frac{1}{z}\right)\sum_{j=0}^{B-1}\binom{T-1}{j}z^{B-j}\tag{6.5}\\
&\,\,\color{blue}{=B\binom{T-1}{B}}
\end{align*}
and the claim (5) follows.
Comment:
In (6.1) we change the order of summation $j\to B-j$.
In (6.2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
In (6.3) we use $\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$.
In (6.4) we shift the index of the middle sum by starting with $j=0$.
In (6.5) we observe the factor in front of the sum is zero. :-)
\begin{align*}
B-(T-B)\frac{1}{z}&=B-(T-B)\left(\frac{T}{B}-1\right)^{-1}\\
&=B-(T-B)\frac{B}{T-B}=0
\end{align*}
