For your form
everything is completely predictable by congruences as long as
$\gcd(a,b) = 1$ and both $a,b$ are squarefree.
Oh, your original form $2 x^2 + 5 y^2 - 10 z^2$ integrally represents all numbers that are neither $4^k (8n+1)$ nor $25^k (5n \pm 1).$ For the negative of the form, switch to $4^k (8n-1),$ so between them there is no $2$-adic obstruction.
I found the most important bit in Watson's book, page 111. Watson defines his discriminant $d$ the same way I do except mine is positive, anyway take the Hessian matrix of second partial derivatives of the function, take its determinant, then divide by two. (for him also negate). For a diagonal ternary $U x^2 + V y^2 + W z^2$ we get $d = 4 UVW.$
For all ternary forms, a genus has class number $1$ whenever $d$ is not divisible by any $p^3$ for odd prime $p,$ or by $64.$
For diagonal form $U x^2 + V y^2 + W z^2,$ class number $1$ if no $p^3$ and $UVW$ is not divisible by $16.$
We get a stronger result if we are told the form is indefinite; an indefinite ternary genus has class number $1$ whenever $d$ is not divisible by any $p^3$ for odd prime $p,$ or by $512.$
For indefinite diagonal form $U x^2 + V y^2 + W z^2,$ class number $1$ if no $p^3$ and $UVW$ is not divisible by $128.$
Therefore the two examples on pages 252-253 of CASSELS illustrate that these bounds are sharp. From Jones and Pall (1939), positive $x^2 + y^2 + 16 z^2$ has another class in the genus, $ 2 x^2 + 2 y^2 + 5 z^2 + 2 y z + 2 z x $ which is spinor regular.
From Siegel about 1951, indefinite $x^2 - 2 y^2 + 64 z^2$ has another class in the genus, $ 4 x^2 - 2 y^2 + 17 z^2 + 4 z x $
Here is an example of mine, I am very proud of it. Of course, it may be published somewhere as well. Two indefinite forms in the same genus, where all the usual properties can be proved by elementary techniques,
$$ x^2 + 100 y^2 - 5 z^2, $$
$$ 4 x^2 + 25 y^2 - 5 z^2. $$
We know these are in the same genus because the binary forms $x^2 + 100 y^2$ and $4 x^2 + 25 y^2$ are in the same genus of positive binaries. We know they are not equivalent because the second one does not integrally represent $1,$ or any $T^2,$ where all prime factors $p$ of $T$ satisfy $p \equiv \pm 1 \pmod 5.$ The second one does represent $q^2,$ with prime $q \equiv \pm 3 \pmod 5.$ The first one represents all squares, but does not primitively represent any $m^2,$ for any $m \equiv \pm 3 \pmod 5.$ These are the three basic properties of spinor exceptions and primitive spinor exceptions, see page 352 in Exceptional Integers for Genera of Integral Ternary Positive Definite Quadratic Forms, Rainer Schulze-Pillot, Duke Mathematical Journal, volume 102, (2000), number 2, pages 351-357. This kind of thing is the same for positive and indefinite ternaries, other things differ; we don't always get elementary proofs (factoring, representing useful primes by certain binary forms) of these simple properties, so examples are nice.