I'm trying to find an answer to this question. Let $K(k)$ be the elliptic integral of the first kind and $K'=K(\sqrt{1-k^2})$. According to Abel's theorem (see this link) we know that if $\frac{K'}{K}=\frac{a+b\sqrt{n}}{c+d\sqrt{n}}$ where $a,b,c,d,n$ are integers, then $k$ is the root of an algebraic equation with integer coefficients. (As I understand it, by algebraic equation with integer coefficients they mean a polynomial with integer coefficients?) So I solved the equation $$ \displaystyle \frac{K(\sqrt{1-k^2})}{K(k)}=\sqrt{2}-1 $$ numerically with 200 digit accuracy and found the corresponding $k$:
$$ k=0.995942004483485267626221391206685115425856878468293704688032877263852669\\384783141641717390148240933985687938743309287701383005201421919573984016706\\406957193055047393475454459046127151355572762924 $$
Here is the Mathematica code I used:
k:=Sqrt[m/.FindRoot[EllipticK[1 - m]/EllipticK[m] - Sqrt[ 2 ] +1==0,{m,1},WorkingPrecision -> 200]]; N[k,200]
(As was pointed out below by ccorn EllipticK(k^2)$=K(k)$.)
Then I tried to identify this algebraic constant by Inverse symbolic calculator but it couldn't identify it. Is there some other way to identify this equation with integer coefficients?
I think Mathworld's explanation of Abel's Theorem is incomplete.
Part I. The Mathematica command is,
$$\frac{K'(k)}{K(k)}=\frac{\text{EllipticK[1-ModularLambda[}\tau]]}{\text{EllipticK[ModularLambda[}\tau]]}=\sqrt{2}-1\tag1$$
where the argument $\tau$ is,
$$\color{brown}{\tau = \sqrt{-2}-\sqrt{-1}}\tag2$$
Thus, we get your constant,
$$k = \sqrt{\lambda(\tau)}=\sqrt{\text{ModularLambda[}\tau]}=0.9959420044834\dots\tag3$$
Other expressions are,
$$\big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$
with the nome $q = \exp(i \pi \tau)$, Dedekind eta function $\eta(\tau)$, and Jacobi theta function $\vartheta_n(0,q)$, with both functions built in Mathematica. See also this post.
Part II. Now as to whether $k$ given by $(3)$ is algebraic, your question is equivalent to asking if the eta quotient,
$$x =\frac{\eta(2\tau)}{\eta(\tau)}\tag4$$
is algebraic. If $\tau$ is an imaginary quadratic $\tau_2$, then it is yes. However, your $\tau$ is an imaginary quartic root $\tau_4$, a root of,
$$\color{brown}{\tau^4+6\tau^2+1 = 0}\tag5$$
Conclusion: I think Abel's Theorem covers only at most $\tau_2$, not $\tau_4$, and Mathworld forgot(?) that crucial detail. Re this comment in an MO post.