$ (a,b), (c,d) \in \mathbb{R} \times \mathbb{R}\ $ let us define a relation by $(a,b) \sim (c,d)$ if and only if $\ a + 2d = c+2b$

That's a correct direct approach. More generally note that $\,(a,b)\sim (c,d)\iff f(a,b) = f(c,d)\,$ for $\,f(u,v) = u-2v\,$ so we can apply the following very general kernel criterion.

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It is straightforward to prove relations of form $\rm\, x\sim y {\overset{\ def}{\color{#c00}\iff}} f(x) = f(y)\, $ are equivalence relations.

More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\color{#c00}\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation $\rm\color{#0a0}{properties\ (E)}\,$ of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f$ as follows

• reflexive $\rm\quad\ \color{#0a0}{\overset{(E)}\Rightarrow}\, f(v) \approx f(v)\:\color{#c00}\Rightarrow\:v\sim v$

• symmetric $\rm\,\ u\sim v\:\color{#c00}\Rightarrow\ f(u) \approx f(v)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(v)\approx f(u)\:\color{#c00}\Rightarrow\:v\sim u$

• transitive $\rm\ \ \ u\sim v,\, v\sim w\:\color{#c00}\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(u)\approx f(w)\:\color{#c00}\Rightarrow u\sim w$

Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.\,$ The equivalence classes $\,f_c = f^{-1}(c)\,$ are called the fibers or preimages, or level sets / curves of $f.$

Yours is the special case when $\,\approx\,$ is the equivalence relation of equality.

You can find many other examples of equivalence kernels in prior answers.

See also the more general notions of difference kernels and equalizers,

Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.