Misprint(s) In "Combinatorial Identities" by Riordan? On page 38 in the exercises I was able to confirm that
$$a_{kj}(x)=(x+1)\dots(x+k-1)a_{k-1,j}(x)+x^{k-1}a_{k-1,j-1}(x+1).$$
For $k\ge{}j$ the author defines $b_{kj}(x)$ via
$$a_{kj}(x)=(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\dots(x+k-1)b_{kj}(x),a_{kk}(x)=b_{kk}(x),$$
where $a_{kk}(x)=x^{k-1}(x+1)^{k-2}\cdots(x+k-2)$.
In both its 1968 and 1979 edition on pages 38 and 39, I am confident that the author intended
$$b_{kj}(x)=(x+1)\dots(x+j)b_{k-1,j}(x)+x^{k-1}b_{k-1,j-1}(x+1)$$
and not
$$b_{kj}(x)=(x+1)\dots(x+j)b_{k-j,j}+x^{k-1}b_{k-1,j-1}(x+1).$$
The absence of the argument $x$ in the second line makes me suspect two typographical errors.
Furthermore, I am confused by the sum that defines $b_j(x,y)$.  A few lines prior to its definition it is stated that $k\ge{j}$ and then its stated that
$$b_j(x,y)=\sum_{k=j}b_{kj}(x)y^{k-j}$$.
The author concludes $b_1(x,y)=[1-(x+1)y]^{-1}(1-xy)^{-1}$ which I suspect arises from
$$[1-(x+1)y](1-xy)b_1(x,y)=1.$$
I am hopeful that somebody has deciphered the possible errors or has some clues how to proceed.  Thank you!
Sincerely,
John Majewicz, Ph.D.
 A: First question: Your assumption is correct. There are two typos as indicated by you.
We have the recurrence relation
\begin{align*}
a_{k,0}(x)&=(x+1)\cdots(x+k-1)a_{k-1,0}(x)\\
&=(x+1)^{k-1}(x+1)^{k-2}\cdots(x+k-1)\\
a_{k,j}(x)&=(x+1)\cdots(x+k-1)a_{k-1,j}(x)\\
&\qquad+x^{k-1}a_{k-1,j-1}(x+1)\qquad\qquad\qquad\qquad 0<j<k\tag{1}\\
a_{k,k}(x)&=x^{k-1}a_{k-1,k-1}(x+1)\\
&=x^{k-1}(x+1)^{k-2}\cdots(x+k-2)\\
a_{1,1}(x)&=a_{0,0}(x)=1
\end{align*}
According to the relation
\begin{align*}
\color{blue}{a_{k,j}(x)}&\color{blue}{=(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}}\\
&\color{blue}{\qquad\cdot\ldots\cdot(x+k-1)b_{k,j}(x)\qquad\qquad\qquad k\geq j}\tag{2}\\
\color{blue}{a_{k,k}(x)}&\color{blue}{=b_{k,k}(x)}
\end{align*}
we derive by putting (2) into (1)
\begin{align*}
&(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k,j}(x)\\
&\qquad=(x+1)\cdots(x+k-1)(x+j+1)^{k-j-2}(x+j+2)^{k-j-3}\\
&\qquad\qquad\cdot\ldots\cdot(x+k-2)b_{k-1,j}(x)\\
&\qquad\qquad+x^{k-1}(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j-1}(x+1)\\
&\qquad=(x+1)\cdots(x+j)(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j}(x)\\
&\qquad\qquad+x^{k-1}(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j-1}(x)\\
\end{align*}

from which after division by $(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)$
\begin{align*}
b_{k,j}(x)&=(x+1)\cdots(x+j)\color{blue}{b_{k-1,j}(x)}\\
&\qquad+x^{k-1}b_{k-1,j-1}(x+1)\tag{3}
\end{align*}
follows.

Second question:
From
\begin{align*}
b_j(x,y)\sum_{k\geq j}b_{k,j}(x)y^{k-j}
\end{align*}
we want to derive $b_1(x,y)$. Since
\begin{align*}
b_1(x,y)=\sum_{k\geq 1}b_{k,1}(x)y^{k-1}\tag{4}
\end{align*}
we start with calculating $b_{k,1}(x)$. We obtain from (3) with $b_{1,1}(x)=1$

\begin{align*}
\color{blue}{b_{k,1}(x)}&=(x+1)b_{k-1,1}(x)+x^{k-1}b_{k-1,0}(x+1)\\
&=(x+1)^2b_{k-2,1}(x)+(x+1)x^{k-2}+x^{k-1}\\
&=(x+1)^3b_{k-3,1}(x)+(x+1)^2x^{k-3}+\cdots+x^{k-1}\\
&\vdots\\
&=(x+1)^{k-1}+(x+1)^{k-2}x+\cdots+x^{k-1}\\
&=\sum_{j=0}^{k-1}(x+1)^{k-1-j}x^j\\
&=(x+1)^{k-1}\sum_{j=0}^{k-1}\left(\frac{x}{x+1}\right)^j\\
&=(x+1)^{k-1}\cdot\frac{1-\left(\frac{x}{x+1}\right)^k}{1-\frac{x}{x+1}}\\
&=(x+1)^k\left(1-\left(\frac{x}{x+1}\right)^k\right)\\
&\,\,\color{blue}{=(x+1)^k-x^k}\tag{5}
\end{align*}

Putting (5) into (4) we obtain

\begin{align*}
\color{blue}{b_1(x,y)}&=\sum_{k\geq 1}b_{k,1}(x)y^{k-1}\\
&=\sum_{k\geq 1}(x+1)^ky^{k-1}-\sum_{k\geq 1}x^ky^{k-1}\\
&=(x+1)\sum_{k\geq 0}\left((x+1)y\right)^k-x\sum_{k\geq 0}(xy)^k\\
&=\frac{x+1}{1-(x+1)y}-\frac{x}{1-xy}\\
&\,\,\color{blue}{=\frac{1}{(1-(x+1)y)(1-xy)}}
\end{align*}
according to the claim.

