These assertions can be proved using (formal) generating functions. Using that for $j\geq 0, k\geq 1$ \begin{align*} \sum_{n\geq 0} {n-j+kj \choose kj} t^n &=\frac{t^j}{(1-t)^{kj+1} }\;\;\mbox{ and }\\ \sum_{n\geq 0}{n+j \choose kj} t^n&=\frac{t^{kj-j}}{(1-t)^{kj+1}}\;\;\;, \end{align*} gives that \begin{align*} \sum_{n\geq 0} t^n F_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt}{(1-t)^k}}\;\;\mbox{ and }\\ \sum_{n\geq 0} t^n G_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt^{k-1}}{(1-t)^k}}\;\;\;, \end{align*}

(I) consider first the (simpler) Gould-Hsu case. Here $c_{n,k,j}= (-1)^{n-j} [t^{n-j} ]\, C_k(t)^{kj+1}$ where $C_k(t)$ denotes the $k$-ary tree function, which is defined by $$C_k(t)=1+tC_k(t)^k\;\;\;.$$ Thus \begin{align*} \sum_{j\geq 0} c_{n,k,j} F_j^{(k)}(x)&=\sum_{j\geq 0} F_j^{(k)}(x)(-1)^{n-j}[t^{n-j}]C_k(t)^{kj+1}\\ &=\sum_{j\geq 0} F_j^{(k)}(x)(-1)^{n-j}[t^{n}] t^j C_k(t)^{kj+1}\\ &=(-1)^n [t^n] \sum_{j\geq 0} F_j^{(k)}(x)(-1)^{j}t^j C_k(t)^{kj+1}\\ &=(-1)^n [t^n] \frac{C_k(t)}{1+tC_k(t)^k}\frac{1}{1+\frac{xtC_k(t)^k}{(1+tC_k(t)^k)}^k}\\ &=(-1)^n [t^n] \frac{1}{1+xt}=x^n\\ \end{align*}

(II) now to your case above. Here $a_{n,k,j}=(-1)^{kj}[t^{n-(k-1)j}] A_{k-1}(t)^{kj+1}$ where $yA_k(y)$ is the inverse of $y(z)=\sum_{j=1}^k (-1)^{j-1} z^j$ . A similar computation as above here gives \begin{align*} \sum_{j\geq 0} a_{n,k,j} F_j^{(k)}((-1)^kx)&=[t^n] \frac{A_{k-1}(t)}{1-T(t)} \frac{1}{1- (-1)^k\frac{T(t)x}{(1-T(t))^k}} \end{align*} where $T(t):=(-1)^kt^{k-1}A_{k-1}(t)^k$. This will simplify to the generating function for the $G_n^{(k)}$ if
$$\frac{A_{k-1}(t)}{1-T(t)}=\frac{1}{1-t}\;\;.$$ And this in turn follows (for $k\geq 2$) with simple stets after substituting $x=tA_{k-1}(t)$ in the equality $$t=\frac{x+(-1)^{k-1}x^k}{1+x}$$.

(III) The Ansatz $b_{n,k,j}=[t^n] T(t)^j Z(t)$ leads to the generating function \begin{align*} \frac{Z(t)}{1-T(t)} \frac{1}{1- (-1)^k\frac{T(t)x}{(1-T(t))^k}} \end{align*} for $R_n(x):=\sum_{j \geq 0} b_{n,k,j} F_j^{(k)}(x)$. One will expect this to be a simple function of $t$ only if $\frac{Z(t)}{1-T(t)}$ and $\frac{T(t)}{(1-T(t)}$ simplify to simple functions of $t$, i.e. can be "solved" for $t$. The targetet generating functions more or less require that $Z=C_k, T=-tC_k^k $ in case (I), resp. that $Z=A_{k-1}, T=t^{k-1}A_{k-1}$ in case (II), this explains the appearance of these special inverse series.

These assertions can be proved using (formal) generating functions. Using that for $j\geq 0, k\geq 1$ \begin{align*} \sum_{n\geq 0} {n-j+kj \choose kj} t^n &=\frac{t^j}{(1-t)^{kj+1} }\;\;\mbox{ and }\\ \sum_{n\geq 0}{n+j \choose kj} t^n&=\frac{t^{kj-j}}{(1-t)^{kj+1}}\;\;\;, \end{align*} gives that \begin{align*} \sum_{n\geq 0} t^n F_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt}{(1-t)^k}}\;\;\mbox{ and }\\ \sum_{n\geq 0} t^n G_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt^{k-1}}{(1-t)^k}}\;\;\;, \end{align*}

(I) consider first the (simpler) Gould-Hsu case. Here $c_{n,k,j}= (-1)^{n-j} [t^{n-j} ]\, C_k(t)^{kj+1}$ where $C_k(t)$ denotes the $k$-ary tree function, which is defined by $$C_k(t)=1+tC_k(t)^k\;\;\;.$$ Thus \begin{align*} \sum_{j\geq 0} c_{n,k,j} F_j^{(k)}(x)&=\sum_{j\geq 0} F_j^{(k)}(x)(-1)^{n-j}[t^{n-j}]C_k(t)^{kj+1}\\ &=\sum_{j\geq 0} F_j^{(k)}(x)(-1)^{n-j}[t^{n}] t^j C_k(t)^{kj+1}\\ &=(-1)^n [t^n] \sum_{j\geq 0} F_j^{(k)}(x)(-1)^{j}t^j C_k(t)^{kj+1}\\ &=(-1)^n [t^n] \frac{C_k(t)}{1+tC_k(t)^k}\frac{1}{1+\frac{xtC_k(t)^k}{(1+tC_k(t)^k)}^k}\\ &=(-1)^n [t^n] \frac{1}{1+xt}=x^n\\ \end{align*}

(II) now to your case above. Here $a_{n,k,j}=(-1)^{kj}[t^{n-(k-1)j}] A_{k-1}(t)^{kj+1}$ where $yA_k(y)$ is the inverse of $y(z)=\sum_{j=1}^k (-1)^{j-1} z^j$ . A similar computation as above here gives \begin{align*} \sum_{j\geq 0} a_{n,k,j} F_j^{(k)}((-1)^kx)&=[t^n] \frac{A_{k-1}(t)}{1-T(t)} \frac{1}{1- (-1)^k\frac{T(t)x}{(1-T(t))^k}} \end{align*} where $T(t):=(-1)^kt^{k-1}A_{k-1}(t)^k$. This will simplify to the generating function for the $G_n^{(k)}$ if
$$\frac{A_{k-1}(t)}{1-T(t)}=\frac{1}{1-t}\;\;.$$ And this in turn follows (for $k\geq 2$) with simple steps after substituting $x=tA_{k-1}(t)$ in the equality $$t=\frac{x+(-1)^{k-1}x^k}{1+x}$$.

(III) The Ansatz $b_{n,k,j}=[t^n] T(t)^j Z(t)$ leads to the generating function \begin{align*} \frac{Z(t)}{1-T(t)} \frac{1}{1- \frac{T(t)x}{(1-T(t))^k}} \end{align*} for $R_n(x):=\sum_{j \geq 0} b_{n,k,j} F_j^{(k)}(x)$. One will expect this to be a simple function of $t$ only if $\frac{Z(t)}{1-T(t)}$ and $\frac{T(t)}{(1-T(t))^k}$ simplify to simple functions of $t$, i.e. can be "solved" for $t$. The targetet generating functions more or less require that $Z=C_k, T=-tC_k^k $ in case (I), resp. that $Z=A_{k-1}, T=-t^{k-1}A_{k-1}$ in case (II), this explains the appearance of these special inverse series.