Is it possible to get all choices of integers satisfying the specified conditions using the families of parameters provided? Suppose $m$,$x$ and $y$ are positive integers such that $\gcd(x,y)=1$ and set  $p=mx$ and $q=my$.
Let $s$ and $r$ be positive integers such that $s$ and $r$ are each relatively prime to both $p$ and $q$ and impose the condition that $s \equiv \pm r^{\pm 1}$ (mod $m$).

The question I am hoping to get answered is the following: Given any choice of $p$ and $q$, can I get every possible choice of $r$ and $s$ that satisfy the conditions specified above using the the two families of integers below.

Family 1
Integers $a,b,c,d$ satisfying $\gcd(a,b,c,d)=\gcd(a^2-c^2,b^2-d^2)=1$ with

*

*b+d=p

*b-d=q

and c+a and c-a satisfy


*$\tau(c+a)=s$


*$\mu(c+a)= r$
where $\tau = (c-a)^{-1} \in \mathbb{Z}_{b+d}$; that is, $\tau$ is the multiplicative inverse of $(c-a)$ modulo $\mathbb{Z}_{b+d}$.
Similarly, $\mu = (c-a)^{-1} \in \mathbb{Z}_{b-d}$.
Family 2
$a,b,c,d$ integers with $\gcd(a,b,c,d)=1$ and $\gcd(a^2-c^2,b^2-d^2)=4$ with
i. $\frac{b+d}{2}=mx$
ii. $\frac{b-d}{2}=my$
and $\frac{c-a}{2}$ and $\frac{c+a}{2}$ satisfy
iii. $\tau\frac{c+a}{2}=s$
iv. $\mu \frac{c-a}{2}=r$
where $\tau=(\frac{c-a}{2})^{-1} \in \mathbb{Z}_{\frac{b+d}{2}}$ and $\mu=(\frac{c-a}{2})^{-1} \in \mathbb{Z}_{\frac{b-d}{2}}$

I know that using family 1, any choice of integers $a,b,c,d$ which satisfy the conditions (1)-(4) are automatically satisfy the $\gcd$ condtions for $a,b,c,d$. I think it is possible to get all values of $p$ and $q$ and $s=r$ using family 1 when $p$ and $q$ have the same parity.

 A: The answer to your question is basically yes, except the $2$ "families" as currently described don't cover all of the cases you requested. In particular, you wrote you want to

... impose the condition that $s \equiv \pm r^{\pm 1} \pmod{m}$.

However, in both of your families, you are defining $s$ and $r$ using a common factor (i.e., $c + a$ or $\frac{c + a}{2}$). Also, you are using a multiplier that is the multiplicative inverse of the same value (i.e., $c - a$ or $\frac{c - a}{2}$), although of different moduli (i.e., $mx$ and $my$), but both moduli have a factor of $m$. This results in $\tau^{-1} \equiv \mu^{-1} \pmod{m} \implies \tau \equiv \mu \pmod{m}$ and, thus, you will only be getting values where
$$s \equiv r \pmod{m} \tag{1}\label{eq1A}$$
This answer assumes this is what you want. I also noticed you seem to acknowledge this limitation with your last sentence, i.e.,

I think it is possible to get all values of $p$ and $q$ and $s=r$ using family 1 when $p$ and $q$ have the same parity.

As you indicated, for any $p$ and $q$ with the same parity, your family $1$ conditions will give you any $s$ and $r$ which meets the stated conditions (except that \eqref{eq1A} is the only relation modulo $m$). First, note your statements $1$ and $2$ mean $2b = p + q \implies b = \frac{p + q}{2}$ and $2d = p - q \implies d = \frac{p - q}{2}$, so they're both integers. Next, have
$$c + a = 1 \implies a = 1 - c \tag{2}\label{eq2A}$$
Thus, $\gcd(a,c) = 1$, so this means $\gcd(a,b,c,d) = 1$, as required. Also, \eqref{eq2A} gives that
$$t_1 = c - a = c - (1 - c) = 2c - 1 \tag{3}\label{eq3A}$$
Thus, varying $c$ means $t_1$ can be any odd integer. You next define
$$\tau \equiv (c-a)^{-1} \equiv t_1^{-1} \pmod{mx} \tag{4}\label{eq4A}$$
Using \eqref{eq2A} gives $s = \tau(c + a) = \tau$, which results in
$$s \equiv t_1^{-1} \pmod{mx} \iff t_1 \equiv s^{-1} = s_1 \pmod{mx} \tag{5}\label{eq5A}$$
Similarly, we also get that
$$r \equiv t_1^{-1} \pmod{my} \iff t_1 \equiv r^{-1} = r_1 \pmod{my} \tag{6}\label{eq6A}$$
The next issue to consider is, due to $t_1$ being restricted to be an odd integer, whether or not there will always be such a solution to \eqref{eq5A} and \eqref{eq6A}. The parities matching mean $p$ and $q$ are either both even or both odd. For the case of both being even, then $s$ and $r$ must be odd (since they're relatively prime to $p$ and $q$). For $t_1$ to be equivalent to the inverses of $r$ and $s$ modulo even values means $t_1$ must be odd.
The other parity case is if both $p$ and $q$ are odd. If we get a result of $t_1$ being even, then just add $\operatorname{lcm}(p, q)$ (which is odd) to it to then get an odd value where both \eqref{eq5A} and \eqref{eq6A} are still satisfied.
Next, using \eqref{eq5A}, there's a $k_1 \in \mathbb{Z}$ such that
$$t_1 = r_1 + k_1(mx) \tag{7}\label{eq7A}$$
Using \eqref{eq6A}, there's a $k_2 \in \mathbb{Z}$ where
$$t_1 = s_1 + k_2(my) \tag{8}\label{eq8A}$$
Thus, \eqref{eq7A} minus \eqref{eq8A} gives
$$\begin{equation}\begin{aligned}
0 & = r_1 + k_1(mx) - (s_1 + k_2(my)) \\
s_1 - r_1 & = m(k_1x - k_2y)
\end{aligned}\end{equation}\tag{9}\label{eq9A}$$
Since $s \equiv r \pmod{m} \implies s_1 \equiv r_1 \pmod{m}$, so using this and \eqref{eq9A} means there's a $k_3 \in \mathbb{Z}$ where
$$s_1 - r_1 = k_{3}(m) \implies k_{3} = k_1x - k_2y \tag{10}\label{eq10A}$$
Since $\gcd(x,y) = 1$, then Bézout's identity states there are integers $d$ and $e$ such that
$$dx + ey = 1 \tag{11}\label{eq11A}$$
Multiplying both sides by $k_{3}$ and using this in \eqref{eq10A} shows that $k_1 = k_3(d)$ and $k_2 = -k_3(e)$. From this in \eqref{eq9A}, we can reverse the steps to get \eqref{eq7A} and \eqref{eq8A} which gives $t_1$, with this then satisfying the requirements of \eqref{eq5A} and \eqref{eq6A}. Finally, we can then get $c$ from \eqref{eq3A} and then $a$ from \eqref{eq2A}.
One final issue to consider is to verify $\gcd(a^2 - c^2, b^2 - d^2) = 1$. Well, $a^2 - c^2 = -t_1$ and $b^2 - d^2 = pq$. Since \eqref{eq5A} shows $\gcd(t_1, p) = 1$ and \eqref{eq6A} shows $\gcd(t_1, q) = 1$, this statement requirement is met.

The other main case to handle is when $p$ and $q$ have opposite parities, for which family $2$ can be used. Here, the i and ii statements gives $b = mx + my$ and $d = mx - my$, so both are odd integers, which means $8 \mid b^2 - d^2$.
Similar to before, have
$$\frac{c + a}{2} = 1 \implies c + a = 2 \implies a = 2 - c \tag{12}\label{eq12A}$$
To avoid $8 \mid a^2 - c^2$ (which would cause the $\gcd$ to be $\gt 4$) requires $a$ (and, thus, also $c$) to be even. With this restriction, we get
$$a^2 - c^2 = (2 - c)^2 - c^2 = 4 - 4c + c^2 - c^2 = 4(1 - c) \tag{13}\label{eq13A}$$
This means $8 \not\mid \gcd(a^2 - c^2, b^2 - d^2)$, so it doesn't contradict the requirement of $\gcd(a^2 - c^2, b^2 - d^2) = 4$. In addition, since $\gcd(a, c) = 2$, plus both $b$ & $d$ are odd, then $\gcd(a, b, c, d) = 1$ is met. Next, also similar to before, we get \eqref{eq12A} giving
$$t_2 = \frac{c - a}{2} = \frac{c - (2 - c)}{2} = \frac{2c - 2}{2} = c - 1 \tag{14}\label{eq14A}$$
Similar to what was determined earlier, since $c$ can be any even integer, this means $t_2$ can be any odd integer. The remaining steps are basically the same as those for the earlier case where $p$ and $q$ had the same parity, so I won't repeat them here. Nonetheless, one point I will mention is that since $t_2$ must be relatively prime to both $p$ and $q$, with one of them being even, then $t_2$ must be odd. Thus, there's no need to possibly adjust the final value due to the procedure resulting in an even value, as was a potential issue earlier.
