According to [Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170](http://link.springer.com/article/10.1007%2FBF01389959)  Theorem 3.11, we can compute
the composition
$BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$
where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer.
The result is that it should be the sum with possibly signs of the identity and
the switch map.


Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$
where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of [Feshbach 
"The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169.](http://www.ams.org/journals/bull/1977-83-03/S0002-9904-1977-14274-2/S0002-9904-1977-14274-2.pdf) So the two signs have to differ.


On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left(
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition
$BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.


[Added afther the comment by Oscar Randal-Williams]


The map $2(id-Switch)$ should be non-trivial for the following reason.


Let's compute the induced maps in homology of infinite loop space by $2id$ 
and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e\_i\otimes e\_j$ the
generator of $H\_i(BO(1))\otimes H\_j(BO(1))$, and we identify it with its image
in $H\_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id)
$ sends, let's say, $e\_4\otimes e\_2$ to $e\_2^2\otimes e\_1^2$, whereas
$\Omega ^{\infty}(2Switch)$ sends $e\_4\otimes e\_2$ to $e\_1^2 \otimes e\_2^2$. 


As $H\_\*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.


[Added Nov.30]
Here is another argument to show that $2id$ 
and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I\_n$-adic completion (probably a simple 
2-adic completion should suffice). Now, the identity and the switch map 
induce different maps on $E^\*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^\*(BO(1)^2)$ as well.


[Added Dec. 07 after a private communication with Oscar Randall-Williams]
(edited Dec 09)


What goes wrong in the above was the assumption that the "coefficients"
are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate
Burnside ring c.f. [Mitchell-Priddy "A double coset formula for levi subgroups and splitting $BGL \_n$"](http://link.springer.com/chapter/10.1007%2FBFb0085237)  This issue is more ore less adressed in
Remark (1.6), as well as in the remark following Theorem (2.3).



Double coset formulas for Orthogonal groups [Solved]
What goes wrong in the above was the assumption that the "coefficients"
are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate
Burnside ring c.f. [Mitchell-Priddy "A double coset formula for levi subgroups and splitting $BGL \_n$"](http://link.springer.com/chapter/10.1007%2FBFb0085237)  This issue is more ore less adressed in
Remark (1.6), as well as in the remark following Theorem (2.3).


