First let's notice that \begin{split} \frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\ &=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \end{split} and similarly $$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$

Hence, the identity in question is equivalent to $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$ being invariant under replacement of $a$ with $2t-a$.

Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$ and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$ Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where: \begin{split} F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\ &= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m}, \end{split} and \begin{split} G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\ &=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}. \end{split}

Using Lagrange–Bürmann formula, we derive the expession $$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$ where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$. It remains to notice that $A-B$ is a multiple of $y$ and in the expansion $$F(x,y) = \left( \frac{A+B}2 \right)^{n/g} \sum_{j=0}^{2m} \binom{n/g}j \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$ the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide. QED

First let's notice that \begin{split} \frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\ &=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \end{split} and similarly $$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$

Hence, the identity in question is equivalent to $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$ being invariant under replacement of $a$ with $2t-a$.

Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$ and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$ Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where: \begin{split} F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\ &= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m}, \end{split} and \begin{split} G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\ &=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}. \end{split}

Using Lagrange–Bürmann formula, we derive the expession $$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$ where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$. It remains to notice that $A-B$ is a multiple of $y$ and in the expansion $$F(x,y) = (-4)^m\left( \frac{A+B}2 \right)^{n/g} \sum_{j=0}^{2m} \binom{n/g}j \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$ the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide. QED

First let's notice that \begin{split} \frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\ &=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \end{split} and similarly $$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$

Hence, the identity in question is equivalent to $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$ being invariant under replacement of $a$ with $2t-a$.

Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$ and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$ Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where: \begin{split} F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\ &= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m}, \end{split} and \begin{split} G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\ &=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}. \end{split}

Using Lagrange–Bürmann formula, we derive the expession $$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$ where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$. It remains to notice that $A-B$ is a multiple of $y$ and in the expansion $$F(x,y) = \left( \frac{A+B}2 \right)^{n/g} \sum_{j=0}^{2m} \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$ the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide. QED

First let's notice that \begin{split} \frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\ &=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \end{split} and similarly $$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$

Hence, the identity in question is equivalent to $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$ being invariant under replacement of $a$ with $2t-a$.

Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$ and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in $$\sum_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$ Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where: \begin{split} F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\ &= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m}, \end{split} and \begin{split} G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\ &=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}. \end{split}

Using Lagrange–Bürmann formula, we derive the expession $$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$ where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$. It remains to notice that $A-B$ is a multiple of $y$ and in the expansion $$F(x,y) = \left( \frac{A+B}2 \right)^{n/g} \sum_{j=0}^{2m} \binom{n/g}j \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$ the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide. QED