(Edit: I revise most of my question as my first answer overlooked that $d_1$ and $d_2$ are odd.)
$\DeclareMathOperator{\res}{res}$
Let $E$ be an elliptic curve over $\mathbb{Q}$ with $E[2]\subset E(\mathbb{Q})$. Let $S$ be the normal $2$-Selmer group and let $S'$ be the relaxed Selmer group, that is the subset of $H^1\bigl(\mathbb{Q},E[2]\bigr)\cong {}^{\mathbb{Q}^\times}\!/{}_{\square} \times{}^{\mathbb{Q}^\times}\!/{}_{\square}$ that satisfy all local conditions away from $2$, but maybe not at $2$.
Within $S'$ define $S''$ the subset of all $\xi=(d,e)$ where both $d$ and $e$ are odd. That is also a sort of a natural Selmer group, where the local condition at $2$ is that $\res_2(\xi)$ lies in the kernel of the valuation ${}^{\mathbb{Q}_2^\times}\!/{}_{\square} \times{}^{\mathbb{Q}_2^\times}\!/{}_{\square}\to {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}\times {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$. The question is: When is $S''\subset S$?
This is not a local question as the elements in the kernel of the valuation in $H^1\bigl(\mathbb{Q}_2,E[2]\bigr)$ are not naturally compared with the image of the local Kummer map $\kappa_2\colon {}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)} \to H^1\bigl(\mathbb{Q}_2,E[2]\bigr)$. Here I give the answer in one particular case; the methods generalise to other similar cases and one could determine the answer in general I imagine. Since the actual calculations with Hilbert symbols in very simple, it may help to understand better the original proof alluded to in the question.
Assume $E(\mathbb{Q}_2)[4]=E(\mathbb{Q}_2)[2]$. That is no point of order $2$ is divisible by $2$ in $E(\mathbb{Q}_2)$.
Also assume that $E$ is given by $y^2=x(x-a)(x-b)$ with $a$ and $b$ odd integers. (I don't think $a\equiv b\pmod{4}$ is needed here.)
First consider the following exact sequence coming from global duality:
$$
0\to S\to S' \xrightarrow{\alpha} H^1\bigl(\mathbb{Q}_2, E\bigr)[2]\xrightarrow{\hat\beta} \hat{S}
$$
where $\alpha$ and $\beta$ are restriction maps.
First if $\beta\colon S\to {}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$ is surjective, then $S'=S$ and hence $S''\subset S$.
Therefore, suppose $\beta$ is not surjective.
My assumption above imposes that the three $2$-torsion points (which are global points and hence in the image of $\beta$) are all distinct in ${}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$. Therefore the image of $\beta$ is equal to group generated by the torsion points. It is $2$-dimensional in the $3$-dimensional target.
Let now $\xi\in S''$ with $\res_2(\xi) = (d,e)$. To say that $\xi\in S$ is equivalent to $\alpha(\xi)=0$. Since we know already that $\xi\in\ker(\hat\beta)$, all we need to check is that $\res_2(\xi)\cup \kappa(P)=0$ for one point $P\in E(\mathbb{Q}_2)$ which is not in the image of $\beta$. Here $\cup \colon H^1\bigl(\mathbb{Q}_2,E[2]\bigr)\times H^1\bigl(\mathbb{Q}_2,E[2]\bigr) \to {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$ is the local duality pairing. Under our identification it corresponds to the pairing $(d,e)\cup (d',e') = (d,e')_2 + (d',e)_2$ on ${}^{\mathbb{Q}_2^\times}\!/{}_{\square} \times{}^{\mathbb{Q}_2^\times}\!/{}_{\square}$ where $(,)_2$ is the Hilbert symbol with values in ${}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$.
In our concrete case, there is a point $P$ with $x$-coordinate equal to $\tfrac{1}{4}$ on $E(\mathbb{Q}_2)$ because $a$ and $b$ are both odd. In fact the points $P$, $T_1=(0,0)$ and $T_2=(a,0)$ generate ${}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$ because $T_2$ has bad reduction, $T_1$ has good non-trivial reduction, and $P$ has trivial reduction. We have $\kappa(P)=(1,1-4a)=(1,-3)$ since $a$ is odd. Now
$$
(d,e)\cup(1,-3) = (d,-3)_2 + (1,e)_2 = 0
$$
because $d$ is odd and $-3\equiv 1 \pmod{4}$. Therefore $(d,e)\in S$.
Finally, I would like to point to Theorem 3 in Appendix 1 of Swinnerton-Dyer's 2-descent through the ages where a question of that nature is discussed, but for places away from 2.
Edit: This is the original answer.
I actually have troubles believing the statement you make in your question. Maybe I am wrong or I misunderstood something, but since it is too long for a comment, I include it as a possible "answer".
Take $y^2 = x^3 - x$. So $n=a=b=1$. I claim: The relaxed Selmer group, imposing all conditions but the one at $2$, is $\mathbb{F}^3$, while the rank of the curve is $0$ so the usual Selmer group is of dimension $2$.
Concretely, the local conditions away from $2$ and $\infty$ impose that $d_1$ and $d_2$ both belong to the group generated by $-1$ and $2$ modulo squares, since the curve has good reduction away from $2$.
The condition at $\infty$ imposes that $d_1$ and $d_2$ have the same sign.
Now we are down to the group generated by $(1,2)$, $(-1,-1)$ and $(2,1)$.
The first two correspond to the torsion points $(1,0)$ and $(0,0)$ in $E(\mathbb{Q})[2]$. So the torsor
$$\begin{align*}
2u^2 -v^2 & = 1 \\ 2u^2 -2w^2 &= -1
\end{align*}$$
is locally soluble at all places, except for $2$.
The reason, I was suspicious initially is that your statement, using global duality, would be equivalent to the surjectivity of the map from the Selmer group $S_2(E/\mathbb{Q})$ to $E(\mathbb{Q}_2)/2E(\mathbb{Q}_2)$. Your question would be equivalent to the surjectivity of the restriction from the Selmer group of some isogeny $\varphi$ to the group of points at this one place. And I don't see a reason why this should hold without conditions on $\varphi$.
