Found it! (Sorry, Doug, ha ha.)
Augustus de Morgan added several appendices to his Elements of Arithmetic in the fifth edition, 1846 (available on Google Books). Appendix 10, pages 201-210, is "on combinations." The relevant paragraph is on 202-203.
Required the number of ways in which a number can be compounded of odd numbers, different orders counting as different ways. If $a$ be the number of ways in which $n$ can be so made, and $b$ the number of ways in which $n+1$ can be made, then $a+b$ must be the number of ways in which $n+2$ can be made; for every way of making $12$ out of odd numbers is either a way of making $10$ with the last number increased by $2$, or a way of making $11$ with a $1$ annexed. Thus, $1+5+3+3$ gives $12$, formed from $1+5+3+1$ giving $10$. But $1+9+1+1$ is formed from $1+9+1$ giving $11$. Consequently, the number of ways of forming $12$ is the sum of the number of ways of forming $10$ and of forming $11$. Now, $1$ can only be formed in $1$ way, and $2$ can only be formed in $1$ way; hence $3$ can only be formed in $1+1$ or $2$ ways, $4$ in only $1+2$ or $3$ ways. If we take the series $1$, $1$, $2$, $3$, $5$, $8$, $13$, $21$, $34$, $55$, $89$, &c. in which each number is the sum of the two preceding, then the $n$th number of this set is the number of ways (orders counting) in which $n$ can be formed of odd numbers. Thus, $10$ can be formed in $55$ ways, $11$ in $89$ ways, &c.
He established "increasing" and "annexing" in deriving the formula for the number of what we now call compositions. He does not treat either of the other two restrictions mentioned above.
Here is a combinatorial proof.
It is more convenient to prove the equivalent formula (obtained by setting $n=m+2j$)
$$\sum_k \binom{k}{j}\binom{m+2j}{2k}=2^{m-1}\frac{m+2j}{m+j}\binom{m+j}{j}.\tag{$*$}$$
To motivate the proof we first prove the closely related but slightly easier
formula
$$\sum_k \binom{k}{j}\binom{m+2j+1}{2k+1} = 2^m\binom{m+j}{j}.\tag{1}$$
Let us first give a generating function proof of $(1)$. We start by multiplying the summand by $x^j y^m$ and summing on $j$ and $m$. We find that
$$
\sum_{j,m}\binom{k}{j}\binom{m+2j+1}{2k+1} x^j y^m = \frac{(x+y^2)^k}{(1-y)^{2k+2}}
$$
and that
$$\sum_{k=0}^\infty \frac{(x+y^2)^k}{(1-y)^{2k+2}}=\frac{1}{1-x-2y} = \sum_{j,m}2^m\binom{m+j}{j}x^j y^m.\tag{2}$$
It is not hard to find objects counted by $2^m\binom{m+j}{j}$. We need to interpret $(x+y^2)^k/(1-y)^{2k+2}$ as counting some of these objects, and this is what the following proof does.
Let $S$ be the set of words in the the letters $x, y_1, y_2$ containing $j$ occurrences of $x$ and a total of $m$ occurrences of $y_1$ and $y_2$. There are $\binom{m+j}{m}$ words with $j$ occurrences of $x$ and $m$ occurrences of $y$. Each $y$ can be replaced with $y_1$ or $y_2$ so there are $2^m\binom{m+j}{m}$ words in $S$.
Now we count the words in $S$ in a different way. Let us define a stopper of a word in $S$ to be an occurrence of either $x$ or $y_2y_1$, and let us call the subwords before, between, and after the stoppers segments, so a word with $k$ stoppers has $k+1$ segments. Thus each segment is of the form $y_1^a y_2^b$, where $a$ and $b$ are nonnegative integers. For example, the word $y_1 y_2 x y_2 y_1 y_2 y_2x$ has three stoppers, $x$, $y_2y_1$, and $x$, and four segments, $y_1y_2$, $\varnothing$, $y_2y_2$, and $\varnothing$. (Here $\varnothing$ denotes the empty word.)
Let us count words in $S$ with $k$ stoppers (and thus $k+1$ segments). We must have $k\ge j$, since every $x$ is a stopper. Then there are $k-j$ stoppers of the form $y_2y_1$ so there are $m-2(k-j)$ occurrences of $y_1$ and $y_2$ among the $k+1$ segments. The segments are of the form $y_1^{a_0}y_2^{b_0},y_1^{a_1}y_2^{b_1},\dots,y_1^{a_k}y_2^{b_k}$, where $a_0+b_0+\cdots +a_k+b_k=m-2(k-j)$. The number of solutions in nonnegative integers of this equation is $\binom{m+2j+1}{2k+1}$
(since for fixed $r$ and $s$ the number of nonnegative integer solutions of $c_1+c_2+\cdots +c_r = s$ is $\binom{r+s-1}{s}=\binom{r+s-1}{s-1}$).
Once the $a_i$ and $b_i$ are determined, the $k$ slots for the stoppers are fixed, and
we may decide which are $x$ and which are $y_2y_1$ in $\binom{k}{j}$ ways. Thus the number of elements of $S$ is
$$
\sum_k \binom{k}{j}\binom{m+2j+1}{2k+1},
$$
which must also be equal to $2^m\binom{m}{j}$.
To prove the original identity $(*)$, we replace $2k+2$ with $2k+1$ in the denominator of the left side of $(2)$. This corresponds to counting only those words in $S$ for which $a_0=0$, i.e., the words in $S$ that do not start with $y_1$. The number of solutions of $a_0+b_0+\cdots +a_k+b_k=m-2(k-j)$ with $a_0=0$ is $\binom{m+2j}{2k}$, so the number of these words is the left side of $(*)$. But (assuming $j$ and $m$ are not both 0) the number of words in $S$ that start with $y_1$ is $2^{m-1}\binom{m+j-1}{j}$, so the number of words in $S$ that do not start with $y_1$ is
$$2^j\binom{m+j}{j} -2^{m-1}\binom{m+j-1}{j} = 2^{m-1}\frac{m+2j}{m+j}\binom{m+j}{j}.$$
Best Answer
I mentioned in my previous answer that the order ideals corresponding to $(s,s+1)$-cores with distinct parts are precisely the subsets of $\{1,2,\dots,s-1\}$ that contain no consecutive elements. This means that the Dyck paths under investigation are all of height 2, with peaks at each element of the order ideal. Since there are no consecutive elements, the paths will always be "bouncing", in the sense that the associated bounce path is the same as the Dyck path itself.
Let's restrict our attention to the Dyck paths with exactly $j$ peaks. There are exactly $\binom{s-j}{j}$ of these. They will all have area $j$, so it remains to prove that the generating function for their bounce statistic is given by $$t^{\binom{s}{2}-j(s-j)}\binom{s-j}{j}_t=t^{\binom{j}{2}+\binom{s-j}{2}}\binom{s-j}{j}_t.$$ To each subset $\{a_1,a_2,\dots,a_j\}$ of $\{1,2,\dots,s-1\}$ we can associate the subset $\{b_1,b_2,\dots,b_j\}$ of $\{0,1,\dots,s-j-1\}$ given by $b_i=a_i-i$. We can check that the bounce statistic of this subset is equal to $\binom{s-j}{2}+b_1+b_2+\cdots+b_j$, therefore the generating function we want is the coefficient of $x^j$ in the expression $$t^{\binom{s-j}{2}}\prod_{i=0}^{s-j-1}(1+t^ix).$$ From here the desired result follows from the q-binomial theorem.