Proof of a Combinatorial Abel Identity
We see that our identity is in fact
$$\sum_{k=0}^n {tk+r\choose k} {tn-tk+s\choose n-k} - \sum_{k=0}^n {tk+r\choose k} {tn-tk+s\choose n-k} \frac{tk}{tk+r} \\ = {tn+r+s\choose n}.$$
While it would be preferable to solve this using formal power series only it appears we need complex variables for this one. With integers $t,r,s \ge 1$ and starting with the first sum we introduce
$${tk+r\choose k} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k+1}} (1+w)^{tk+r} \; dw$$
and
$${tn-tk+s\choose n-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{tn-tk+s} \; dz.$$
This last integral vanishes when $k\gt n$ so we may extend the sum to infinity, getting
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+s}}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{r}}{w} \sum_{k\ge 0} z^k (1+z)^{-tk} \frac{1}{w^k} (1+w)^{tk} \; dw \; dz.$$
Now with $\epsilon$ and $\gamma$ small in a neighborhood of the origin we get that for this to converge we must have $\epsilon/(1-\epsilon)^t \lt \gamma/(1+\gamma)^t.$ We shall see that we may solve this with an additional constraint, namely that $\gamma \gt\epsilon.$ Doing the summation we find
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+s}}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{r}}{w} \frac{1}{1-z(1+w)^t/w/(1+z)^t} \; dw \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+s}}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} (1+w)^{r} \frac{1}{w-z(1+w)^t/(1+z)^t} \; dw \; dz.$$
The pole at $w=0$ has been canceled. There is a pole at $w=z$ however and with the chosen parameters it is inside the contour. We get for the residue
$$\left.(1+w)^r \frac{1}{1-tz(1+w)^{t-1}/(1+z)^t}\right|_{w=z} = (1+z)^r \frac{1}{1-tz/(1+z)}$$
The derivative would have vanished if the pole had not been simple. Substituting into the outer integral we get
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+r+s+1}}{z^{n+1}} \frac{1}{1-(t-1)z} \; dz.$$
Continuing with the second sum we obtain
$$\sum_{k=1}^n {tk+r\choose k} {tn-tk+s\choose n-k} \frac{tk}{tk+r} = t \sum_{k=1}^n {tk+r-1\choose k-1} {tn-tk+s\choose n-k} \\ = t \sum_{k=0}^{n-1} {tk+t+r-1\choose k} {t(n-1)-tk+s\choose (n-1)-k}.$$
We can recycle the earlier computation and find
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{t(n-1)+t+r-1+s+1}}{z^{n}} \frac{t}{1-(t-1)z} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+r+s}}{z^{n+1}} \frac{tz}{1-(t-1)z} \; dz.$$
Subtracting the two the result is
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+r+s}}{z^{n+1}} \frac{(1+z)-tz}{1-(t-1)z} \; dz = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{tn+r+s}}{z^{n+1}} \; dz.$$
This evaluates to
$${tn+r+s\choose n}$$
by inspection and we have proved the theorem.
To show that the pole at $w=z$ is the only one inside the contour apply Rouche's theorem to
$$h(w) = w(1+z)^t - z(1+w)^t$$
with $f(w) = w (1+z)^t$ and $g(w) = z (1+w)^t.$ We need $|g(w)| < |f(w)|$ on $|w|=\gamma$ and since $f(w)$ has only one root there so does $h(w)$, which must be $w=z.$ We thus require
$$|g(w)| \le |z| (1+\gamma)^t \lt \gamma |1+z|^t = |f(w)|.$$
Now $\gamma/(1+\gamma)^t$ starts at zero and is increasing since $(1+\gamma-\gamma t)/(1+\gamma)^{t+1}$ is positive for $\gamma \lt 1/(t-1)$ with a local maximum there. Since $|z|/|1+z|^t \le \epsilon / (1-\epsilon)^t$ we may choose $\epsilon$ for this to take on a value from the range of $\gamma/(1+\gamma)^t$ on $[0, 1/(t-1)].$ Instantiating $\gamma$ to the right of this point yields a value $\gt \epsilon$ that fulfils the requirements of the theorem. Here we have used that $\epsilon/(1+\epsilon)^t \lt \epsilon/(1-\epsilon)^t \lt \gamma/(1+\gamma)^t$ by construction. No need for Rouche when $t=1.$
Here is a solution more in line with Aigner's hints. Much of this is lifted directly from Knuth's Convolution Polynomials, available on the arXiv.
You were trying to use $(1)$ with $p_n(x)=\binom{tn+x}{n}$, but it turns out the correct method is to use $(2)$ with $$p_n(x)=\binom{tn+x}{n}\frac{x}{x+tn}.$$The result is $$ (x+y)\sum k\binom{tk+x}{k}\frac{x}{x+tk}\binom{t(n-k)+y}{n-k}\frac{y}{y+t(n-k)}=nx\binom{tn+x+y}{n}\frac{x+y}{x+y+tn} $$ Canceling the $x$ and $x+y$, and applying the absorption identities $\binom{tn+x+y}{n}=\frac{tn+x+y}{n}\binom{tn+x+y-1}{n-1}$, and $\binom{tk+x}{k}=\frac{tk+x}{k}\binom{tk+x-1}{k-1}$, we get $$ \sum_k \binom{tk+x-1}{k-1}\binom{t(n-k)+y}{n-k}\frac{y}{y+t(n-k)}=\binom{tn+x+y-1}{n-1} $$ Finally, the result follows by replacing $n$ with $n+1$, reversing the order of summation ($k\leftarrow n+1-k $), and replacing $x$ with $x-t+1$.
Of course, you still need to find a function $F(z)$ for which $$F(z)^x=\sum_{n\ge0}p_n(x)z^n=\sum_{n\ge0}\binom{tn+x}{n}\frac{x}{tn+x}z^n\tag{*}.$$ It turns out that the answer is $$F(z)=\sum_{n\ge0}\binom{tn+1}{n}\frac{z^n}{tn+1}\tag{**}$$ This is a function which satisfies $$ F(z) = 1+zF(z)^t\tag{***} $$ You can use take (***) as a definition of $F$, and recover (**) via Lagrange inversion. Knuth gives an interesting combinatorial proof of how (**) implies (*) in Concrete Mathematics, section 7.5. I think there should be a way to show (***) implies (*) via Lagrange inversion, but so far I have been unsuccessful.