EDIT: I was forgetting two examples with differing discriminants: it is well known, and easy, to show that $x^2 + xy+y^2$ and $x^2 + 3 y^2$ represent exactly the same numbers. With indefinite forms, $x^2 + xy-y^2$ and $x^2 -5 y^2$ represent exactly the same numbers.
ORIGINAL: You seem to be restricting to positive forms. If two forms have the same discriminant and represent the same prime numbers, they are either equivalent or "opposite." Here equivalent would mean $\det M = 1,$ opposite would mean $\det M = -1.$
Next, what about differing discriminants? There are a finite number of pairs of forms of different discriminants that represent the same primes, or the same odd primes, or wjhat have you. I and Kaplansky wrote an informal article on this. Hendrik Lenstra's student John Voight, now at Dartmouth, finished the job. Item number 6 at VOIGHT.
Finishing your question is then a matter of checking pairs of forms from Voight's tables and seeing how they compare on composite numbers. My guess is that perfect agreement is impossible for differing discriminants.
In comparison, for positive forms in three variables, there are infinitely many pairs of forms, of differing discriminants, that represent the same numbers. However, not with the same multiplicities, this is a difficult result due to Alexander Schiemann. I looked into the matter further a year ago, I think I found all pairs of positive ternaries representing the same numbers. No proof of completeness, though.
Schiemann also found an example in four variables, two distinct forms that represent the same numbers with the same multiplicities. It's in the book by Nipp. I believe the discriminant was 1729. Yes, that's correct, 1729.
A binary quadratic form, with integer coefficients, is some
$$ f(x,y) = A x^2 + B xy + C y^2. $$ The discriminant is
$$ \Delta = B^2 - 4 A C. $$
We will abbreviate this by $$ \langle A,B,C \rangle. $$ It is primitive if $\gcd(A,B,C)=1. $
Standard fact, hard to discover but easy to check: $$ (A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$
where $ X = x z - D yw, \; Y = A xw + C yz + B yw. $ This gives us Dirichlet's definition of "composition" of quadratic forms of the same discriminant,
$$ \langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle. $$ In particular, if this $D=1,$ the result represents $1$ and is ($SL_2 \mathbf Z$) equivalent to the "principal" form for this discriminant. Oh, duplication or squaring in the group; if $\gcd(A,B)=1,$
$$ \langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle. $$
This comes up with positive forms: $ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle $ is principal, the group identity.
Probably should display some $SL_2 \mathbf Z$ equivalence rules, these are how we calculate when things are not quite right for Dirichlet's rule:
$$ \langle A,B,C \rangle \cong \langle C,-B,A \rangle, $$
$$ \langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle, $$
$$ \langle A,B,C \rangle \cong \langle A, B - 2 A, A - B +C \rangle. $$
Imaginary first. Suppose we want to know about $\mathbf Q(\sqrt {-47}).$ Reduced positive forms $ \langle A,B,C \rangle $ obey $|B| \leq A \leq C$ and $B \neq -A,$ also whenever $A=C$ we have $B \geq 0.$ Our group of binary forms is
-47
class number 5
all
( 1, 1, 12)
( 2, -1, 6)
( 2, 1, 6)
( 3, -1, 4)
( 3, 1, 4)
This is an abelian group in any case, so it is cyclic of order 5. These are also the five elements in the ring of integers of $\mathbf Q(\sqrt {-47}).$
Here is the mapping from forms to ideals: given $ \langle A,B,C \rangle, $ drop the letter $C.$ That's it.
$$ \langle A,B,C \rangle \mapsto \left[ A, \frac{B + \sqrt \Delta}{2} \right]. $$
Oh, why is this an ideal, rather than just some $\mathbf Z$-lattice? Because, given $\alpha,\beta$ rational integers, $$ \left[ \alpha, \frac{\beta + \sqrt \Delta}{2} \right] $$ is an ideal if and only if
$$ 4 \alpha | ( \Delta - \beta^2 ). $$
Group: we already see how to do
$$ \langle 2,1,6 \rangle^2 \cong \langle 4,1,3 \rangle \cong \langle 3,-1,4 \rangle; $$
$$ \langle 2,1,6 \rangle \circ \langle 3,-1,4 \rangle \cong \langle 2,5,9 \rangle \circ \langle 3,5,6 \rangle \cong \langle 6,5,3 \rangle \cong \langle 3,-5,6 \rangle \cong \langle 3,1,4 \rangle; $$
$$ \langle 2,1,6 \rangle \circ \langle 3,1,4 \rangle \cong \langle 6,1,2 \rangle \cong \langle 2,-1,6 \rangle. $$
$$ \langle 2,1,6 \rangle \circ \langle 2,-1,6 \rangle \cong \langle 1,1,12 \rangle $$
in any case.
Best Answer
As you mentioned on Chat, Kap and I wrote a note on forms of different discriminants (but positive definite forms, meaning negative discriminants). This was corrected and extended by John Voight, now at Dartmouth; also published.
The best known examples are the pair $x^2 + xy + y^2$ and $x^2 + 3 y^2.$ The proof that these represent the same numbers is some 2 by 2 matrices, some things mod 2. Same for the indefinite pair $x^2 + xy - y^2$ and $x^2 - 5 y^2.$
Probably worth pointing out that the forms $x^2 + xy + 2ky^2$ and $x^2 + (8k-1)y^2$ represent all the same odd numbers, including any odd primes. The latter form does not represent $2$ or $-2,$ if you can say the same about the former form they agree on primes. We called these "Trivial Pairs." Um; as with Gauss, we discard these if the discriminant is square, meaning we demand $8k - 1 \neq -w^2,$ or $k \neq \frac{1 - w^2}{8}.$
The question changes if you allow square discriminants.
There may be infinitely many other indefinite pairs, we did not check.
If the discriminant is not a square, two forms of the same discriminant that share even a single prime are $GL_2 \mathbb Z$ equivalent. In traditional terms, they are either equivalent or opposite.
Forms with square discriminant, such as $xy$ or $x^2 - y^2,$ are unusual in representing entire arithmetic progressions. Primes do not control things.
For self study, I recommend Buell, Binary Quadratic Forms. I find it easier reading than Buchmann and Vollmer. I also recommend L. E. Dickson Introduction to the Theory of Numbers. For just the first section, I also like Cox, Primes of the Form $x^2 + n y^2.$ Cox does a good job on positive forms, genera, composition. No indefinite forms, though, no Pell. As you can see from my answers, I like the first chapter in Conway, The Sensual Quadratic Form. The wonderful thing there is the "Topograph" construction. I have written a bunch of software to tell me how to avoid arithmetic mistakes in drawing those. These give the best way for talking about a fixed indefinite form $A x^2 + B x y + C y^2$ with $B^2 - 4 AC > 0 $ but not a square. The "cycle" method of Lagrange does not do well when $|n|$ is too large, in finding all solutions to $A x^2 + B x y + C y^2 = n.$ Lagrange's method gives all answers when $|n| < \frac{1}{2} \sqrt{B^2 - 4 AC};$ this result is Theorem 85 in Dickson. Oh, both Lagrange and Conway are talking about primitive representations, $\gcd(x,y) = 1.$