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Here is a set of generators of a subgroup $H$ of index $2$, such that your group equals $H\times\{-Id\}$ (matrices act on the right):

(Edit : in a previous version, the matrices were orthogonal for another (equivalent) quadratic form.)

\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

\begin{bmatrix} -39 & -152 & 380 \\ -40 & -151 & 380 \\ -20 & -76 & 191 \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 \\ 0 & -9 & 20 \\ 0 & -4 & 9 \\ \end{bmatrix}

\begin{bmatrix} -39 & -38 & 190 \\ -10 & -11 & 50 \\ -10 & -10 & 49 \\ \end{bmatrix}

\begin{bmatrix} -210 & -76 & 931 \\ -140 & -49 & 620 \\ -79 & -28 & 350 \\ \end{bmatrix}

\begin{bmatrix} -170 & 0 & 741 \\ 0 & 1 & 0 \\ -39 & 0 & 170 \\ \end{bmatrix}

\begin{bmatrix} -609 & -1216 & 3800 \\ -320 & -641 & 2000 \\ -200 & -400 & 1249 \\ \end{bmatrix}

\begin{bmatrix} -930 & -2128 & 6251 \\ -560 & -1279 & 3760 \\ -329 & -752 & 2210 \\ \end{bmatrix}

\begin{bmatrix} -1608009 & -3459026 & 10438030 \\ -910270 & -1958101 & 5908810 \\ -549370 & -1181762 & 3566111 \\ \end{bmatrix}

\begin{bmatrix} -23751 & -50350 & 152950 \\ -13250 & -28091 & 85330 \\ -8050 & -17066 & 51841 \\ \end{bmatrix}

\begin{bmatrix} -85689 & -183578 & 554990 \\ -46910 & -100501 & 303830 \\ -28750 & -61594 & 186209 \\ \end{bmatrix}

(The sixth and twelth matrices have infinite order, the other have order 2).

In fact this group $H$ acts on the upper half plane (just as $SL_2(Z)$ does), and there is an equivariant retraction on a tree. The quotient contains exactly one cycle. This action furnishes the above generators.

(Note : it follows the abelianized group has the form $\mathbf Z\oplus$ 2-torsion.)

I'm quite in a hurry and have to go now. Tell me if you need more. (The methods will soon be available on the arxiv).

Edit : this is a new answer, after more computations.

Let $H$ be the subgroup of your orthogonal group that preserve globally each connected component of the (two-sheeted) space $q(x,y,z)=-1$.

Up to this action, there is a single isometry class of isotropic vectors.

One representant is $(-1\ -3\ 8)$ and its stabilizer is the infinite dihedral group generated by

B1 = [ -39 -152  380]   C1 = [-2889 -8398 22610]
     [ -40 -151  380]        [-2210 -6421 17290]
     [ -20  -76  191]        [-1190 -3458  9311]

The group $H$ can be described as follows : let us write

A1 = [-1  0  0]             B1 = [ -39 -152  380]             B6=[ -609 -1216  3800]
     [ 0  1  0]                  [ -40 -151  380]                [ -320  -641  2000
     [ 0  0  1]                  [ -20  -76  191]                [ -200  -400  1249]

A2 = [ 1  0  0]             B2 = [-23751 -50350 152950]       T=[-609  -76 2660]
     [ 0 -1  0]                  [-13250 -28091 85330]          [-380  -49 1660]
     [ 0  0  1]                  [ -8050 -17066 51841]          [-220  -28  961]

A3 = [ 1  0  0]             B3 = [-1608009 -3459026 10438030]
     [ 0 -9 20]                  [ -910270 -1958101  5908810]
     [ 0 -4  9]                  [ -549370 -1181762  3566111]

A4 = [-170    0  741]           B4 = [-194561 -415872 1258560]
     [   0    1    0]                [-109440 -233929  707940]
     [ -39    0  170]                [ -66240 -141588  428489]

A5 = [ -930 -2128  6251]        B5 = [-39 -38 190]
     [ -560 -1279  3760]             [-10 -11  50]
     [ -329  -752  2210]             [-10 -10  49]

All the matrices $Ai$ and $Bi$ have order 2, while $T$ has infinite order.

Let $K$ be the free product of all the subgroups generated by these element (thus it is isomorphic to $\mathbf Z/2^{\star 11}\star \mathbf Z$.

Let $R$ be the subgroup of K generated by $[A1,A2]$ , $[A1,A3]$, $[A3,B1]$, $[A2,A4]$, $[A5,B3]$, and $T.A4.T^{-1}.A5$.

Then $R$ is the kernel of the obvious representation of $K$ in $H$, which is an epimorphism.

The group $\mathrm H_1(H,\mathbf Q)$ has dimension 1, thus $H$ is not a reflective group. I hope to be able to draw a Coxeter Diagram for its reflection subgroup soon. If this happens, I will edit this answer once more.

Here is a set of generators of a subgroup $H$ of index $2$, such that your group equals $H\times\{-Id\}$ (matrices act on the right):

[  49   10   10]
[ -50  -11  -10]
[-190  -38  -39],


[ 350   28   79]
[-620  -49 -140]
[-931  -76 -210],

[ 1249   400   200]
[-2000  -641  -320]
[-3800 -1216  -609],


[  51841   17066    8050]
[ -85330  -28091  -13250]
[-152950  -50350  -23751],

[ 186209   61594   28750]
[-303830 -100501  -46910]
[-554990 -183578  -85689],

In fact this group $H$ acts on the upper half plane (just as $SL_2(Z)$ does), and there is an equivariant retraction on a tree. The quotient contains exactly one cycle. This action furnishes the above generators.

(Note : it follows the abelianized group has the form $\mathbf Z\oplus$ 2-torsion.)

I'm quite in a hurry and have to go now. Tell me if you need more. (The methods will soon be available on the arxiv).

Here is a set of generators of a subgroup $H$ of index $2$, such that your group equals $H\times\{-Id\}$ (matrices act on the right):

(Edit : in a previous version, the matrices were orthogonal for another (equivalent) quadratic form.)

\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

\begin{bmatrix} -39 & -152 & 380 \\ -40 & -151 & 380 \\ -20 & -76 & 191 \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 \\ 0 & -9 & 20 \\ 0 & -4 & 9 \\ \end{bmatrix}

\begin{bmatrix} -39 & -38 & 190 \\ -10 & -11 & 50 \\ -10 & -10 & 49 \\ \end{bmatrix}

\begin{bmatrix} -210 & -76 & 931 \\ -140 & -49 & 620 \\ -79 & -28 & 350 \\ \end{bmatrix}

\begin{bmatrix} -170 & 0 & 741 \\ 0 & 1 & 0 \\ -39 & 0 & 170 \\ \end{bmatrix}

\begin{bmatrix} -609 & -1216 & 3800 \\ -320 & -641 & 2000 \\ -200 & -400 & 1249 \\ \end{bmatrix}

\begin{bmatrix} -930 & -2128 & 6251 \\ -560 & -1279 & 3760 \\ -329 & -752 & 2210 \\ \end{bmatrix}

\begin{bmatrix} -1608009 & -3459026 & 10438030 \\ -910270 & -1958101 & 5908810 \\ -549370 & -1181762 & 3566111 \\ \end{bmatrix}

\begin{bmatrix} -23751 & -50350 & 152950 \\ -13250 & -28091 & 85330 \\ -8050 & -17066 & 51841 \\ \end{bmatrix}

\begin{bmatrix} -85689 & -183578 & 554990 \\ -46910 & -100501 & 303830 \\ -28750 & -61594 & 186209 \\ \end{bmatrix}

(The sixth and twelth matrices have infinite order, the other have order 2).

In fact this group $H$ acts on the upper half plane (just as $SL_2(Z)$ does), and there is an equivariant retraction on a tree. The quotient contains exactly one cycle. This action furnishes the above generators.

(Note : it follows the abelianized group has the form $\mathbf Z\oplus$ 2-torsion.)

I'm quite in a hurry and have to go now. Tell me if you need more. (The methods will soon be available on the arxiv).

Here is a set of generators of a subgroup $H$ of index $2$, such that your group equals $H\times\{-Id\}$ :

[  49   10   10]
[ -50  -11  -10]
[-190  -38  -39],


[ 350   28   79]
[-620  -49 -140]
[-931  -76 -210],

[ 1249   400   200]
[-2000  -641  -320]
[-3800 -1216  -609],


[  51841   17066    8050]
[ -85330  -28091  -13250]
[-152950  -50350  -23751],

[ 186209   61594   28750]
[-303830 -100501  -46910]
[-554990 -183578  -85689],

In fact this group $H$ acts on the upper half plane (just as $SL_2(Z)$ does), and there is an equivariant retraction on a tree. The quotient contains exactly one cycle. This action furnishes the above generators.

(Note : it follows the abelianized group has the form $\mathbf Z\oplus$ 2-torsion.)

I'm quite in a hurry and have to go now. Tell me if you need more. (The methods will soon be available on the arxiv).

Here is a set of generators of a subgroup $H$ of index $2$, such that your group equals $H\times\{-Id\}$ (matrices act on the right):

[  49   10   10]
[ -50  -11  -10]
[-190  -38  -39],


[ 350   28   79]
[-620  -49 -140]
[-931  -76 -210],

[ 1249   400   200]
[-2000  -641  -320]
[-3800 -1216  -609],


[  51841   17066    8050]
[ -85330  -28091  -13250]
[-152950  -50350  -23751],

[ 186209   61594   28750]
[-303830 -100501  -46910]
[-554990 -183578  -85689],

In fact this group $H$ acts on the upper half plane (just as $SL_2(Z)$ does), and there is an equivariant retraction on a tree. The quotient contains exactly one cycle. This action furnishes the above generators.

(Note : it follows the abelianized group has the form $\mathbf Z\oplus$ 2-torsion.)

I'm quite in a hurry and have to go now. Tell me if you need more. (The methods will soon be available on the arxiv).