In general, there is no good way to find the conjugate of a composition. But you can use the following fact:
Fact. Let $M$ be a linear subspace of $\mathbb R^n$, and let $g:M\to\mathbb R\cup\{+\infty\} $ be a convex function. If $\pi_M$ is the orthogonal projection onto $M$, then
$$
(g\circ \pi_M)^*(y) = \begin{cases} g^*(y),\quad &y\in M \\ +\infty,\quad &y\notin M \end{cases} \tag1
$$
Proof. Let $M^\perp$ be the orthogonal complement of $M$. Every vector $x$ decomposes as $x_M+x_{M^\perp}$ where $x_M\in M$ and $x_{M^\perp}\in {M^\perp}$. Therefore,
$$(g\circ \pi_M)^*(y) =\sup_{x_M, \ x_{M^\perp}} \left(\langle y, x_M\rangle+ \langle y, x_{M^\perp}\rangle - f(x_M) \right) \tag2
$$
Observe that $x_{M^\perp}$ appears only in the term $\langle y, x_{M^\perp}\rangle$. If $y\in M$, then $\langle y, x_{M^\perp}\rangle=0$ for all $x_{M^\perp}$ and therefore (2) becomes the formula for $g^*$. If $y\notin M$, then $\sup_{ x_{M^\perp}} \langle y, x_{M^\perp}\rangle =+\infty$. $\Box$
Combining the above fact with your knowledge of one-dimensional conjugate
$$|cx|^*(y)=\begin{cases} 0 , \quad &|y|\le |c| \\ +\infty , \quad &\text{otherwise} \end{cases} \tag3$$
you can conclude with
$$|\langle a,x\rangle|^*(y)=\begin{cases} 0 , \quad &y=ta, \ -1\le t\le 1 \\ +\infty , \quad &\text{otherwise} \end{cases} \tag4$$
$$\sup_{\frac{x}{t}\in dom~f\\ t>0}(t(\dfrac{y^Tx}{t}+s-f(\dfrac{x}{t}))) = \sup\{tv \mid v =\sup_{u\in dom~f}((y^Tu+s-f(u)))\text{ and }\color{red}{t > 0}\}$$
Edit to address a question in the comments
What does $$\sup_{\frac{x}{t}\in dom~f\\ t>0}(t(\dfrac{y^Tx}{t}+s-f(\dfrac{x}{t})))$$ mean?
It means we form a set $$A =\left\{t(\dfrac{y^Tx}{t}+s-f(\dfrac{x}{t}))~ \right|~\left. t > 0, \frac xt \in dom~f\right\}$$
Then we find its supremum.
But that is a bit ungainly. After all, everywhere that $x$ occurs in the definition of $A$, it is as a part of $x/t$. So why not just use that value directly instead of expressing it in terms of this other arbitrary value $x$? So I replaced $x/t$ with $u$ to get
$$A = \left\{t(y^Tu+s-f(u))\mid t > 0, u \in dom~f\right\}$$
For the next step, I just made use of this equivalence:
$$\sup\{\phi(r,s)\mid r\in A, s\in B\} = \sup\{\sup\{\phi(r,s)\mid s\in B\}\mid r \in A\}$$
where $\phi : A\times B \to \Bbb R$. This is easy enough to see: For any value $p$ in the LH set, $p = \phi(r,s)$ for some $r,s$, and so $p\le \sup\{\phi(r,s)\mid s\in B\} \le \sup\{\sup\{\phi(r,s)\mid s\in B\}\mid r \in A\}$. Thus the LH side is $\le$ the RH side. Conversely, For any value $h$ less than the RH side, there must be an $r \in A$ with $\sup\{\phi(r,s)\mid s\in B\} > h$, which in turn means there is an $s \in B$ with $\phi(r,s) > h$, which means that the LH side is also $> h$. Hence they must be equal.
Of course this leaves you with a problem, which wasn't apparent in my original answer, because I wrote down the wrong thing. You will note that originally I had $tv \in dom~f$. I had meant this to be $t > 0$ (which correction is now highlighted in red, if your browser shows colored mathjax) but in my various editings of the form, I had accidently left in the wrong thing and deleted the correct one.
But in this form, it is quite clear. Since $t$ can be arbirarily large, the common value of these two expressions is $\infty$ (assuming that $f^*$ has at least one positive value). Once it is made obvious by my version, you can check that it holds in your version as well. Pick an arbitrary $u \in dom~f$ with $f(u) > 0$ and for any $t > 0$, you can take $x = tu$ to make the supremum as large as you like.
There are two possibilities here. Either the point of this exercise is for you to discover the complex conjugate of a perspective function is always infinite, or else, by conjugate of the perspective function, they only mean to conjugate with respect to constant $t$:
$$ g^*(y,t)=\sup_{\frac{x}{t}\in dom~f}(t(\dfrac{y^Tx}{t}+s-f(\dfrac{x}{t})))$$
I can't say which is the case for your course.
Best Answer
I don't know if there is an easier approach, using some theorems on the convex conjugate maybe, but below I am just working with the definitions you gave: If we write $$\langle x,y\rangle - f(x) = \langle y - \frac{A}{2}x + b, x\rangle$$
Then we see, that we just have to show that for a fixed $y$ the maximum is attained at $$x = A^{-1} (b+y)$$
Thus, we fix $y$ and differentiate
$$\frac{\partial}{\partial x} \langle y - \frac{A}{2}x + b, x\rangle = y - \frac{A}{2} x + b - \frac{A}{2} x $$
Setting this to 0 will yield the solution as required, provided that A is positive definite.