Here's "why this works", and also how to find a formula (with exhaustive search, essentially) when one exists. But really, the answer is that no such formula may exist in general; we're just lucky it exists for this particular set of 7 values mod 19.
The basic idea is that for any integer $k$ relatively prime to $19$, the function $y \mapsto ky$ modulo 19 is invertible. In other words, instead of looking at a set of integers $y$ modulo 19, you can look at the corresponding set of integers $ky$ mod 19, and if that set happens to have a simple characterisation for some $k$, it amounts to a characterisation of your original set of values.
So consider a table of $ky \bmod 19$, for $y$ in your list $[0, 3, 6, 8, 11, 14, 17]$, and integers $k$. Only integers $k$ from $1$ to $18$ need be considered, as $(k+19)y \equiv ky \mod 19$. No need to consider $k=0$ as $0$ is not relatively prime to $19$: $0y \bmod 19$ is always $0$, so it doesn't tell you anything about $y$. Now it so happens that for $k=7$, the values modulo 19 that are attained happen to be consecutive modulo 19. For instance, this Python code:
for k in range(1,19): values = [(k*y)%19 for y in [0, 3, 6, 8, 11, 14, 17]] attained = ['*' if x in values else '.' for x in range(19)] print "%2d: %-30s %s" % (k, values, ''.join(attained))
prints
1: [0, 3, 6, 8, 11, 14, 17] *..*..*.*..*..*..*. 2: [0, 6, 12, 16, 3, 9, 15] *..*..*..*..*..**.. 3: [0, 9, 18, 5, 14, 4, 13] *...**...*...**...* 4: [0, 12, 5, 13, 6, 18, 11] *....**....***....* 5: [0, 15, 11, 2, 17, 13, 9] *.*......*.*.*.*.*. 6: [0, 18, 17, 10, 9, 8, 7] *......****......** 7: [0, 2, 4, 18, 1, 3, 5] ******............* 8: [0, 5, 10, 7, 12, 17, 3] *..*.*.*..*.*....*. 9: [0, 8, 16, 15, 4, 12, 1] **..*...*...*..**.. 10: [0, 11, 3, 4, 15, 7, 18] *..**..*...*...*..* 11: [0, 14, 9, 12, 7, 2, 16] *.*....*.*..*.*.*.. 12: [0, 17, 15, 1, 18, 16, 14] **............***** 13: [0, 1, 2, 9, 10, 11, 12] ***......****...... 14: [0, 4, 8, 17, 2, 6, 10] *.*.*.*.*.*......*. 15: [0, 7, 14, 6, 13, 1, 8] **....***....**.... 16: [0, 10, 1, 14, 5, 15, 6] **...**...*...**... 17: [0, 13, 7, 3, 16, 10, 4] *..**..*..*..*..*.. 18: [0, 16, 13, 11, 8, 5, 2] *.*..*..*..*.*..*..
Now look at each row in the picture on the right, until you notice that for $k=7$, all the asterisks are consecutive (cyclically) — the values taken are $[18, 0, 1, 2, 3, 4, 5]$. You can add 1 to shift it cyclically, then the values will be $[0, 1, 2, 3, 4, 5, 6]$. Thus the set of values $(7y+1) \bmod 19$ for $y$ in your original list is the above, which has a nice simple characterisation as the set of nonnegative integers less than 7. Using this characterisation gives the formula $(7y + 1) \bmod 19 < 7$. (If you look at the table you see that $k=12$ also has the values consecutive; this is not surprising because $12 \equiv -7 \mod 19$.)
We just happened to get lucky. Starting with a different set of 7 integers may not give any such formula. For instance if we change "11" to "12", we get the picture:
1: [0, 3, 6, 8, 12, 14, 17] *..*..*.*...*.*..*. 2: [0, 6, 12, 16, 5, 9, 15] *....**..*..*..**.. 3: [0, 9, 18, 5, 17, 4, 13] *...**...*...*...** 4: [0, 12, 5, 13, 10, 18, 11] *....*....****....* 5: [0, 15, 11, 2, 3, 13, 9] *.**.....*.*.*.*... 6: [0, 18, 17, 10, 15, 8, 7] *......**.*....*.** 7: [0, 2, 4, 18, 8, 3, 5] *.****..*.........* 8: [0, 5, 10, 7, 1, 17, 3] **.*.*.*..*......*. 9: [0, 8, 16, 15, 13, 12, 1] **......*...**.**..
(I've omitted $k=10$ to $k=18$) in which the asterisks are not consecutive for any $k$. This doesn't mean that no clever formula of any sort can exist, but it does mean that no formula of the form $ky + l < m$ will work for any triple $(k, l, m)$.
It may be an interesting probability exercise to calculate how lucky we got. :-) (Given a modulus like $19$, what is the probability that for a set of integers of a certain size, for some $k$ the multiples are consecutive?)