We begin by proving the claims made by Blue in the edit to the question.
Let $I$ be the incentre of $ABC$, and let $R$ be its circumradius. Since $O$ and $H$ lie inside $ABC$, the triangle must be acute.
The incircle is divided into three arcs by its points of contact with the sides of $ABC$. At least one of these arcs, say the one nearest vertex $A$, contains neither $O$ nor $H$. Thus when the rays $AO$ and $AH$ meet the incircle at $O$ and $H$, respectively, each of these rays is intersecting the incircle for the second time. Moreover, as in any triangle, $AI$ bisects angle $OAH$. It follows that the points $O$ and $H$ are symmetric about $AI$. In particular, $AH = AO = R$.
In any triangle, $\overrightarrow{AH} = 2\overrightarrow{OA'}$, where $A'$ is the midpoint of $BC$. Hence $OA' = R/2$. It follows from this that $\angle BOC = 120^{\circ}$, hence that $\angle BAC = 60^{\circ}$.
If we introduce $O'$ as in the figure (the reflection of $O$ through $A'$), then $O$, $B$ and $C$ belong to the circle with radius $R$ centred at $O'$. Since $\overrightarrow{AH} = \overrightarrow{OO'}$, the quadrilateral $OAHO'$ is a rhombus with side $R$. Consequently, $H$ also belongs to circle $BOC$.
If $J$ is the point halfway along arc $OH$ on circle $BOC$, then $BJ$ bisects $\angle OBH$, hence $J$ lies on $BI$. Similarly, $J$ lies on $CI$. Hence $J = I$, and $I$ lies on circle $BOC$. Since $AI$ bisects $\angle BAC$, it meets the circumcircle of $ABC$ again at $O'$, which is midway between $B$ and $C$.
Conversely, we carry out a construction corresponding to the above requirements. Start with a circle centred at $O$ with radius $1$. Mark two points $B$ and $C$ on the circle so that $\angle BOC = 120^{\circ}$. Let $O'$ be the reflection of $O$ through $BC$. Then $O'$ is on the circle. Now let $I$ be any point on circle $BOC$, on the same side of $BC$ as $O$. (We will specify $I$ further below.) Let $O'I$ cut $BO'C$ again at $A$. Let $H$ be the reflection of $O$ through $O'I$. Then reversing the arguments above, we find that $H$ is the orthocentre and $I$ the incentre of triangle $ABC$, and that $IH = IO$.
The only question that remains is how to choose $I$ on circle $BOC$ so that $IO$ is equal to the inradius of $ABC$. If we let $x$ be the inradius, then $x$ is the distance from $I$ to line $BC$. We also have $IO^2 = (1/2 - x)^2 + 1 - (x+1/2)^2 = 1-2x$. The condition $OI = x$ is equivalent to $x^2 = 1 - 2x$, or $x = \sqrt{2} - 1$.
Thus the construction can be completed by letting $I$ be a point of intersection of circle $BOC$ with a circle centred at $O$ with radius $\sqrt{2}-1$.
I'm not sure how to motivate this last step geometrically.
Summary of my construction Given two points $O$ and $O'$, write $R = OO'$. Construct the circles $K$ and $K'$ of radius $R$ centred at $O$ and $O'$, respectively. Let $B$ and $C$ be the points of intersection. Let $I$ be a point of intersection of $K'$ with the circle of radius $(\sqrt{2}-1)R$ centred at $O$. Then let $A$ be the point of intersection of $O'I$ with $K$.
Alternative construction (using $IA = 2IO$, proved by dxiv below) Instead of constructing $I$, construct $A$ directly by intersecting $K$ with the circle of radius $(2\sqrt{2} - 1)R$ centred at $O'$.
Summary of dxiv's construction Construct a triangle $AIO$ with $IO= r$, $IA = 2r$, $OA = (\sqrt{2}+1)r$. Let $K$ be the circle centred at $O$ passing through $A$. Construct angles of $30^{\circ}$ on either side of $AI$. Let $B$ and $C$ be the intersections with $K$ of the outer sides of these angles.
Here's a construction, where I'll use $s$ for the given hypotenuse:
Construct right $\triangle OAB$ with $|\overline{OA}| = |\overline{OB}| = s$.
Construct $\overline{AC}$ perpendicular to $\overline{AB}$, with $|\overline{AC}| = r/2$.
Construct $D$ where $\overleftrightarrow{BD}$ meets the "far side" the the circle about $C$ through $A$.
Construct $E$, the midpoint of $\overline{BD}$.
Construct $F$, where the circle about $B$ through $E$ meets a semicircle on $\overline{OB}$.
Construct $G$, the reflection of $O$ across $\overline{BF}$ (which is easy, as $\angle OFB$ is necessarily a right angle).
Construct $H$ where $\overline{BG}$ meets the semicircle on $\overline{OB}$.
$\triangle OBH$ is the desired triangle.
Certainly, $\overline{BF}$ bisects angle $B$. That $|\overline{BX}| = r$ is trickier to establish: by the Power of a Point theorem, we have $$|\overline{OX}||\overline{XH}| = |\overline{BX}||\overline{XF}|$$
A bit of messy algebra shows that
$$|\overline{BF}| = \frac{1}{4}\left(\; r + \sqrt{r^2 + 8s^2} \;\right) = \frac{1}{2}\left(\;\frac{r}{2} + \sqrt{\left(\;\frac{r}{2}\;\right)^2 + \left(\;s\sqrt{2}\;\right)^2\;}\;\right)$$
where the right-most expression gives the form that guided the construction. Then, the bisector has the correct length by virtue of the fact that the $\triangle OBH$ is, in fact, the solution.
I suspect there's a construction that makes all the relations clear.
Best Answer
Let $ABG$ be the triangle you want to construct, $F$ its incenter. In the circle $\Gamma$ with diameter $AB$, let $D$ be the endpoint of the diameter perpendicular to $AB$ and on the opposite side of $AB$ to $G$.
Since $AGB$ is a right triangle, $G$ lies on $\Gamma$. Because $GF$ bisects $\angle AGB$, it meets $\Gamma$ at $D$. Furthermore, it is easy to see that $\angle AFB = 135^{\circ}$. It follows that $F$ is on the circle $\Gamma'$ centered at $D$ passing through $A$ and $B$.
Hence $G$ can be constructed as follows. First construct the point $D$. Then let $F$ be the intersection of $\Gamma'$ with the perpendicular to $AB$ through $C$. Lastly, let $G$ be the other intersection of $DF$ with $\Gamma$.