Your generating function will have the form $$f(x)=\sum_{n\ge 0}a_nx^n\;,$$ where $a_n$ is the number of ways to make a total of $n$ dollars using the prescribed coin and bills. Each $\$1$ coin must therefore add $1$ to the exponent, as must each $\$1$ bill; each $\$2$ bill must add $2$ to the exponent, and each $\$5$ bill must add $5$. Thus,
$$f(x)=\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ coins}}\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ bills}}\underbrace{(1+x^2+x^4+\ldots)}_{\$2\text{ bills}}\underbrace{(1+x^5+x^{10}+\ldots)}_{\$5\text{ bills}}\;,$$
or more compactly,
$$f(x)=\left(\sum_{n\ge 0}x^n\right)^2\left(\sum_{n\ge 0}x^{2n}\right)\left(\sum_{n\ge 0}x^{5n}\right)\;.$$
Now use the basic geometric generating function to rewrite this as
$$f(x)=\left(\frac1{1-x}\right)^2\left(\frac1{1-x^2}\right)\left(\frac1{1-x^5}\right)=\frac1{(1-x)^2(1-x^2)(1-x^5)}\;,$$
which can be further simplified to $$f(x)=\frac1{(1-x)^3(1+x)(1-x^5)}\;,$$ if you wish.
The answer to the question is now the coefficient of $x^{100}$ in $f(x)$.
Added: Suppose that you want the number of $\$5$ bills to be at least $2$ and at most $6$. Then the factor that accounts for the $\$5$ bills would be
$$\begin{align*}
x^{2\cdot5}+x^{3\cdot5}+x^{4\cdot5}+x^{5\cdot5}+x^{6\cdot5}&=x^{10}+x^{15}+x^{20}+x^{25}+x^{30}\\
&=x^{10}\left(1+x^5+x^{10}+x^{15}+x^{20}\right)\\
&=\frac{x^{10}\left(1-x^{25}\right)}{1-x^5}\;.
\end{align*}$$
For pennies you have the sequence $\langle 1,1,1,\dots\rangle$, which as you say correponds to the formal power series
$$1+x+x^2+\ldots=\sum_{n\ge 0}x^n\;.$$
That’s just a geometric series, so you know what the generating function is (even if you didn’t realize it right away:
$$\sum_{n\ge 0}x^n=\frac1{1-x}\;.$$
For nickels you can do almost the same thing. Your sequence is $\langle 1,0,0,0,0,1,0,\dots\rangle$, corresponding to another geometric series:
$$1+x^5+x^{10}+\ldots=\sum_{n\ge 0}x^{5n}=\sum_{n\ge 0}\left(x^5\right)^n=\frac1{1-x^5}\;.$$
For pennies and nickels you want
$$\left(1+x+x^2+\ldots\right)\left(1+x^5+x^{10}+\ldots\right)\;:\tag{1}$$
$(1)$ has one $x^n$ term for every way to write $n$ in the form $5m+k$ with $m,k\in\Bbb N$, which is precisely the number of ways to make the amount $n$ with nickels and pennies. Clearly, then, the generating function is just
$$\frac1{(1-x)(1-x^5)}\;.$$
I’ll leave the last part to you.
Best Answer
First observe that
$$\sum_{a+b+2c=k}(-1)^{b+c}\binom{65+a}a\binom{15+b}b\binom{15+c}c$$
is the coefficient of $x^k$ in the product
$$\left(\sum_{a\ge 0}\binom{65+a}ax^a\right)\left(\sum_{b\ge 0}(-1)^b\binom{15+b}bx^b\right)\left(\sum_{c\ge 0}(-1)^c\binom{15+c}cx^{2c}\right)\,.$$
In closed form this is the product
$$\begin{align*} \frac1{(1-x)^{66}}\cdot\frac1{(1+x)^{16}}\cdot\frac1{(1+x^2)^{16}}&=\frac1{(1-x)^{66}(1+x)^{16}(1+x^2)^{16}}\\ &=\frac1{(1-x)^{50}(1-x^2)^{16}(1+x^2)^{16}}\\ &=\frac1{(1-x)^{50}(1-x^4)^{16}}\\ &=\frac1{(1-x)^{50}}\cdot\frac1{(1-x^4)^{16}}\,, \end{align*}$$
which in turn is the closed form of
$$\left(\sum_a\binom{49+a}ax^a\right)\left(\sum_b\binom{15+b}bx^{4b}\right)\,.$$
The coefficient of $x^k$ in this product is
$$\sum_{a+4b=k}\binom{49+a}a\binom{15+b}b\,;$$
can you see why this is the number of ways to make change for $25k$ cents using these $66$ types of coins?