One point of view is, once you have abstracted the "real-world" quantities (number of coins, number of cents) into equations with named variables, you can go at it with algebra and never mind what the next equation "means" in "reality."
You may have to take this view some day for some problem.
For this problem, in the equation $9d+24q=400,$ you know that every coin contributes at least one cent to the total value; the $9$ represents the number of extra cents that each dime contributes, and the $24$ represents the number of extra cents that each quarter contributes.
If the $100$ coins were all pennies we would be $400$ cents short of five dollars, so the $d$ dimes and $q$ quarters need to contribute a total of $400$ "extra" cents:
$$9d+24q=400.$$
Coefficient extraction:
We have
\begin{align*}
D(z)&=\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)\left(1-z^{25}\right)\left(1-z^{50}\right)}
\end{align*}
We know from this answer
\begin{align*}
A(z)&=\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)}\\
&=\sum_{m=0}^\infty\left(\frac{1}{4}\left\lfloor\frac{m}{5}\right\rfloor^2+\frac{5}{4}\left\lfloor\frac{m}{5}\right\rfloor
-\frac{1}{2}\left\lfloor\frac{m}{10}+\frac{1}{2}\right\rfloor+1\right)z^m\tag{1}
\end{align*}
We calculate analogously
\begin{align*}
\color{blue}{B(z)}&=\frac{1}{\left(1-z^{25}\right)\left(1-z^{50}\right)}\\
&=\sum_{q=0}^\infty z^{25q}\sum_{f=0}^\infty z^{50f}\\
&=\sum_{n=0}^\infty\left(\sum_{{25q+50f=n}\atop{q,f\geq 0}}\right)z^n\\
&=\sum_{n=0}^\infty\left(\sum_{{2q+f=n}\atop{q,f\geq 0}}\right)z^{25n}\\
&=\sum_{n=0}^\infty\left(\sum_{q=0}^{\lfloor n/2\rfloor}1\right)z^{25n}\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\left\lfloor \frac{n}{2}\right\rfloor+1\right)z^{25n}}\tag{2}
\end{align*}
Using the coefficient of operator $[z^t]$ to denote the coefficient of $z^t$ of a series we obtain from (1) and (2)
\begin{align*}
\color{blue}{[z^t]}&\color{blue}{D(z)}=[z^t]A(z)B(z)=[z^t]\sum_{m=0}^\infty a_mz^m\sum_{n=0}^\infty b_nz^{25n}\\
&=[z^t]\sum_{q=0}^\infty\left(\sum_{{m+25n=q}\atop{m,n\geq 0}}a_mb_n\right)z^q\\
&=[z^t]\sum_{q=0}^\infty\left(\sum_{n=0}^{\lfloor q/25\rfloor}a_{q-25n}b_n\right)z^q\\
&=\sum_{n=0}^{\lfloor t/25\rfloor}a_{t-25n}b_n\\
&=\sum_{n=0}^{\lfloor t/25\rfloor}
\left(\frac{1}{4}\left\lfloor\frac{t-25n}{5}\right\rfloor^2+\frac{5}{4}\left\lfloor\frac{t-25n}{5}\right\rfloor
-\frac{1}{2}\left\lfloor\frac{t-25n}{10}+\frac{1}{2}\right\rfloor+1\right)\\
&\qquad\qquad\quad\cdot
\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\\
&\,\,\color{blue}{=\sum_{n=0}^{\lfloor t/25\rfloor}
\left(\frac{1}{4}\left(\left\lfloor\frac{t}{5}\right\rfloor-5n\right)^2+\frac{5}{4}\left(\left\lfloor\frac{t}{5}\right\rfloor-5n\right)
-\frac{1}{2}\left\lfloor\frac{t}{10}-\frac{5n}{2}+\frac{1}{2}\right\rfloor+1\right)}\\
&\qquad\qquad\quad\color{blue}{\cdot
\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)}\tag{3}
\end{align*}
It seems, the result (3) can be considerably simplified, since we obtain with the help of Wolfram alpha the nice representation
\begin{align*}
D(z)&=\color{blue}{1} + z + z^2 + z^3 + z^4\\
&\qquad + \color{blue}{2} z^5 + 2 z^6 + 2 z^7 + 2 z^8 + 2 z^9 \\
&\qquad+ \color{blue}{4} z^{10} + 4 z^{11} + 4 z^{12} + 4 z^{13} + 4 z^{14}\\
&\qquad+ \color{blue}{6 }z^{15} + 6 z^{16} + 6 z^{17} + 6 z^{18} + 6 z^{19}\\
&\qquad + \color{blue}{9} z^{20} + 9 z^{21} + 9 z^{22}+ 9 z^{23} + 9 z^{24}\\
&\qquad + \color{blue}{13}z^{25} + 13 z^{26} + 13 z^{27} + 13 z^{28} + 13 z^{29}\\
&\qquad + \color{blue}{18} z^{30} + 18 z^{31} + 18 z^{32} + 18 z^{33} + 18 z^{34}\\
&\qquad+ \color{blue}{24} z^{35} + 24 z^{36}+ 24 z^{37} + 24 z^{38} + 24 z^{39}\\
&\qquad + \color{blue}{31} z^{40} + 31 z^{41} + 31 z^{42} + 31 z^{43} + 31 z^{44}\\
&\qquad+ \color{blue}{39} z^{45} + 39 z^{46 }+ 39 z^{47} + 39 z^{48} + 39 z^{49}\\
&\qquad + O(z^{50})
\end{align*}
with equal coefficients in groups of five.
First-order Asymptotics:
The asymptotic estimation of OP is correct. We find in chapter IV: Complex Analysis, Rational and Meromorphic Asymptotics of Analytic Combinatorics by P. Flajolet and R. Sedgewick the
Proposition IV.2: Let $T$ be a finite set of integers without a common divisor ($\gcd(T) = 1$). The number of partitions with summands restricted to $T$ satisfies
\begin{align*}
P_t^{T}\sim\frac{1}{\tau}\,\frac{t^{r-1}}{(r-1)!},\qquad \text{ with }\tau:=\prod_{\omega\in T}\omega,\qquad r:= \mathrm{card}(T)
\end{align*}
Here we consider
\begin{align*}
[z^t]D(z)=[z^t]\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)\left(1-z^{25}\right)\left(1-z^{50}\right)}\tag{4}
\end{align*}
For first-order asymptotic of coefficients in (4) only the pole at $z=1$, which is the nearest pole to $0$ with highest order $5$ needs to be considered.
We have according to (4) $T=\{1,5,10,25,50\},\tau=1\cdot5\cdot10\cdot25\cdot50,r=4$ from which
\begin{align*}
\color{blue}{[z^t]D(z)\sim} \frac{1}{1\cdot5\cdot10\cdot25\cdot50}\,\frac{t^4}{4!}=\color{blue}{\frac{2}{3}10^{-6}t^4}
\end{align*}
follows.
Chapter 4 of the book provides all the necessary information to derive this asymptotic estimation.
Best Answer
Elimination is a good idea. I think the details are not done correctly. In any case, I would rather eliminate $p$, the arithmetic is easier. We get $24q+9d=399$, or equivalently $$8q+3d=133.$$
We can use general techniques to solve this Diophantine equation, but the numbers are so small that it does not seem worthwhile. Note that when $q=2$, then $133-8q$ is a multiple of $3$, and we get $q=2$, $d=39$, and therefore $p=59$.
Now we have found one solution. We need others.
Go back to the equation $8q+3d=133$. We have found one solution, $q=2$, $d=39$. For any integer $k$, we therefore have $$8(2+3k)+3(39-8k)=133.$$ (Note the cancellation.)
Substitute various values of $k$. With $k=1$, for example, we get the solution $q=5$, $d=31$. That gives $p=64$.
Continue, using $k=2$, $3$, and so on. For the sake of reality, we will have to make sure that our values of $q$, $d$, $p$ are all $\ge 0$. We cannot use $k\gt 4$, for that would give negative $d$.