Failure of existence of GCD

Sure: just build the ring with generators and relations to have the properties you want. For instance, consider the commutative ring $R$ which is generated by elements $a$ and $b$ and infinitely many elements $d_i,s_i,t_i$ with relations that $d_is_i=a$ and $d_it_i=b$ for all $i$. Explicitly, we can realize this $R$ as a subring of the field of rational functions $\mathbb{Q}(a,b,d_i)$ by mapping $s_i$ to $a/d_i$ and $t_i$ to $b/d_i$ (in particular, this shows $R$ is a domain). Even more explicitly, $R$ is the set of rational functions of the form $p/q$ where $q$ is a monomial in the $d_i$ of degree $n$ and $p\in\mathbb{Z}[a,b,d_i]$ is in the ideal $(a,b)^n$. This description makes it easy to verify that all the $d_i$ are maximal common divisors of $a$ and $b$ (since, for instance, if $p/q$ divides $a$ then $p$ can only be $\pm 1$ or $\pm a$ times a monomial in the $d_i$).

You can build a similar example where there are no maximal common divisors: take the subring $R$ of the field of rational functions $\mathbb{Q}(a,b,d_n)$ generated by $a,b,d_n,a/d_n,b/d_n,$ and $d_{n+1}/d_n$ for each $n$. In particular, writing $s_n=a/d_n$, $t_n=b/d_n$, and $u_n=d_{n+1}/d_n$, note that for any $N$, the subring $R_N$ generated by $d_0,s_N,t_N$, and the $u_n$ for $n<N$ is actually freely generated by $d_0, s_N,t_N$, and the $u_n$, and that every finite subset of $R$ is contained in some $R_N$. This makes it easy to verify that $a$ and $b$ have no maximal common divisor in $R$, since their greatest common divisor in $R_N$ is $d_N$ but $d_N$ properly divides $d_{N+1}$ for each $N$.