Suppose for the sake of contradiction $a+c$ is a prime. Notice that $LHS$ is natural so indeed $ab>cd \iff \frac{a}{d}>\frac{c}{b}>1$ so $a>d$. Notice $$b+d\equiv a(b+d)\pmod {a+c}$$So $a+c\mid (a-1)(b+d)$ and since $a+c>a-1$ yields $a+c\mid b+d$. Thus $$b+d\ge a+c>b+d$$Contradiction. So $a+c$ is composite.

Note that \[b+d\equiv a+b+c+d = ab-cd \equiv ab + ad = a(b+d) \pmod{a+c}\]Thus, $(a-1)(b+d) \equiv 0 \pmod{a+c}$. AFTSOC that $a+c$ is prime, this means that $a+c\mid b+d$ because $a-1<a+c$. Write $b+d= k(a+c)$. Then, \[ab-cd = ab - c(k(a+c) - b) = ab +bc - ck(a+c) = (a+c)(b-ck)\]Thus, $k+1 = b-ck$, so $b=ck+k+1$ and $d= ak-k-1$. This describes all solutions, and cannot occur for $c>b$ because \[b= ck+k+1 \geq c+2\]so we have a contradiction, and $a+c$ isn't prime. $\blacksquare$.

Suppose for the sake of contradiction, that $a+c=p\in\mathbb P$. Observe that $a+b+c+d=ab-cd$ is equivalent to the following: $$(a-1)(b-1)=(c+1)(d+1).$$Thus if $\gcd(a-1,c+1)=1$, we would get that $c+1\mid b-1$, which is absurd as $c>b$. Thus, $\gcd(a-1,c+1)=d>1$. Hence, $d\mid (a-1)+(c+1)=a+c=p\implies d=p$. This is absurd as we get that $p=a+c\geq (p+1)+(p-1)=2p$.