After yesterday's adventures in brute force computation, I've settled down to do some theory.
There are in fact arbitrarily large sets of positive integers with the desired
property. Below is a pseudocode algorithm that, given any such set, produces a
larger one. I've not written out the justifications for the steps, but the
verifications are routine except for Step 2, discussed below. The key observation
is that the property is preserved if the set is translated by a common multiple of all
the pairwise differences.
Algorithm
Input: A set of $t > 0$ positive integers $a_1,\ldots,a_t$ with
$\operatorname{gcd}(a_i, a_j) = {\mid} a_i - a_j \mid$ for $i \ne j$.
Output: A set of $t + 1$ positive integers with the same property.
Step 1. (Initialize.)
Set $m = \operatorname{lcm}(\{a_i - a_j\})$ (the least common multiple of all the
pairwise differences), $M = \operatorname{lcm}(\{a_i\})$.
Step 2. (Choose candidate for new entry.)
Find a positive integer $N$ such that the quotients
$q_i := {\mid} N - a_i {\mid}/\operatorname{gcd}(N,a_i)$ are all
coprime to $M$ and to each other.
Step 3. (Done?)
If all the $q_i = 1$, set $a_{t+1} := N$ and return $a_1,...,a_{t+1}$.
Step 4. (Deal with one $q_i$.)
If, for some $j$, $q_j \ne 1$, find a suitable multiple of $m$, say $km$,
so that $q_j$ divides $a_j + km$ (and thus also $N + km$).
Set $a_i := a_i + km$ for each $i$, $N := N + km$, $m := \operatorname{lcm}(m, q_j)$.
Recompute the $q_i := {\mid} N - a_i {\mid}/\operatorname{gcd}(N,a_i)$.
Step 5. (Iterate.)
Go to step $3$.
In Step $2$ it is not obvious that such an $N$ must exist, so here is a formula:
Let $N = M \operatorname{rad}(M)$, where $\operatorname{rad}(M)$ is the radical
of $M$ (that is, the product of the distinct prime factors of $M$). This will work
except in the trivial case $t = 1$, $a_1 = 1$, when $N = 2$ will do. In practice,
this formula gives a very large $N$ and results in large values for the $q_i$,
so searching for something smaller that also works is sensible.
Example
Start with $\{a_i\} = \{6, 8, 9, 12\}$.
We have $m = 24$, $M = 72$.
If we use $N = M \operatorname{rad}(M) = 432$, the $q_i$ are $71, 53, 47, 35$.
As promised, they are pairwise coprime and also coprime to $72$.
For illustration, $N = 16$ is a more convenient choice; the $q_i$ are then
$5, 1, 7, 1$.
To fix $q_1 = 5$, we need to find $k$ so that $5$ divides $24k + 6$.
$k = 1$ will work. The new $a_i$ are $30, 32, 33, 36$; $N$ is now $40$;
$m$ is now $120$; the $q_i$ are now $1, 1, 7, 1$.
Next we need to find $k$ so that $7$ divides $120k + 33$. $k = 2$ works.
The new $a_i$ are $270, 272, 273, 276$; $N$ is $280$ and can now serve as $a_5$.
The result is $270, 272, 273, 276, 280$.