Determining a quadrilateral up to similarity with four angles

There can be more than one possible quadrilateral, given your four angles. I don't have a proof (yet), but found a counterexample with GeoGebra, see figures below.

EDIT.

Here's a way to find all the possible solutions and the lengths of the sides. Let's name the given angles and an auxiliary angle: $$ \angle BAC=\alpha,\quad \angle CBD=\beta,\quad \angle DCA=\gamma,\quad \angle ADB=\delta,\quad \angle ABD=\theta. $$ Suppose now $AB=1$. From the sine rule we get: $$ BC={\sin\alpha\over\sin(\alpha+\beta+\theta)},\quad CD=BC\cdot{\sin\beta\over\sin(\alpha-\gamma+\theta)},\quad DA=CD\cdot{\sin\gamma\over\sin(\alpha+\delta+\theta)} $$ and $$ AB=1=AD\cdot{\sin\delta\over\sin\theta}= {\sin\alpha\sin\beta\sin\gamma\sin\delta\over \sin(\alpha+\beta+\theta)\sin(\alpha-\gamma+\theta) \sin(\alpha+\delta+\theta)\sin\theta}. $$ This equation can be rewritten as: $$ \sin(\alpha+\beta+\theta)\sin(\alpha-\gamma+\theta) \sin(\alpha+\delta+\theta)\sin\theta= \sin\alpha\sin\beta\sin\gamma\sin\delta\cdot (\sin^2\theta+\cos^2\theta)^2 $$ which becomes, after expansion, an omogeneous quartic equation for $\sin\theta$ and $\cos\theta$. Dividing that by $\cos^4\theta$ we thus obtain a quartic algebraic equation for $\tan\theta$. Note however that the original equation has the spurious solution $\theta=-\alpha$. That solution can be factored out, leaving a cubic equation: $$ \begin{align} &\tan^3\theta [\sin\gamma \sin (2 \alpha+\beta+\delta)+2 \cos\gamma \cos(\alpha+\beta) \cos (\alpha+\delta)-\sin\beta \sin\gamma \cos\delta-\cos\beta \sin\gamma \sin\delta]\\ +&\tan^2\theta [2 \sin 2\alpha \sin\gamma \sin(\beta+\delta) +2\cos\gamma \sin(\alpha+\beta)\cos (\alpha+\delta) -2 \cos2\alpha \sin\gamma \cos(\beta+\delta)\\ &\quad\quad+2\cos\gamma\cos(\alpha+\beta)\sin(\alpha+\delta)-2\sin\beta\sin\gamma\sin \delta]\\ &+\tan\theta [-\sin\gamma \sin (2\alpha+\beta+\delta) +2 \cos\gamma \sin(\alpha+\beta) \sin(\alpha+\delta) -\sin\beta\sin\gamma\cos\delta\\ &\quad\quad-\cos\beta\sin\gamma \sin\delta] -2 \sin\beta\sin\gamma\sin\delta \end{align} $$

As an example, setting (as in the figure): $$ \alpha=20°,\quad \beta=30°,\quad \gamma=40°,\quad \delta=50° $$ I used Mathematica to solve the above equation and obtained three real solutions: $$ \theta=30°,\quad \theta=100°,\quad \theta=150°. $$ The last solution isn't acceptable, because of the request that all inner angles in the figure be less than $180°$, which in particular entails $$ \alpha+\beta+\theta<180°,\quad \alpha+\delta+\theta<180°. $$

In this answer we prove that the quadrilaterals found by @Intelligenti pauca have the desired properties.

Consider a convex quadrilateral $ABCD$ with $\angle DBA = 30^\circ$, $\angle CBD=30^\circ$, $\angle BDC = 10^\circ$, $\angle ADB = 50^\circ$. Let $E$ be the reflection of $A$ in $BD$. Note that $E$ lies on $BC$ (since $\angle EBD = \angle DBA = 30^\circ = \angle CBD$) and that $ABE$ is an equilateral triangle (since $AB=BE$ and $\angle EBA=60^\circ$). Since $\angle AEC = 60^\circ = \angle ADC$, and $D$, $E$ lie on the same side of $AC$, the points $A$, $D$, $E$, $C$ are concyclic. But $BD$ is the perpendicular bisector of $AE$, hence $AD=ED$, and therefore the inscribed angles on $AD$ and $ED$ are equal. This shows that $\angle DCA = \angle EAD = 40^\circ$. At this point it is immediate to find $\angle BAC = 20^\circ$.

Now, consider a convex quadrilateral $ABCD$ with $\angle ABD = 100^\circ$, $\angle CBD = 30^\circ$, $\angle CDB = 80^\circ$, and $\angle BDA = 50^\circ$. Let $E$ be the reflection of $B$ in $AD$. Note that $E$ lies on $CD$, since $\angle EDB = 2\angle ADB = 2\cdot 50^\circ = 180^\circ - \angle BDC$. Moreover, $\angle DBE = 90^\circ - \angle ADB = 40^\circ$, hence $\angle EBA = \angle DBA - \angle DBE = 100^\circ - 40^\circ = 60^\circ$. Of course, $AB=AE$, hence $ABE$ is equilateral. In particular, $EB=EA$. Now, observe that $\angle DCB = 70^\circ = \angle CBE$. This shows that $ECB$ is isosceles with $EC=EB$ and $\angle BEC=40^\circ$. Since $EC=EB=EA$, the points $C$, $B$, $A$ lie on a circle with center $E$. As such, $\angle BAC = \frac 12 \angle BEC = \frac 12 \cdot 40^\circ = 20^\circ$. It is now easily calculated that $\angle DCA = 40^\circ$.

Edit: the answer here means there was an error in my original argument. I've rewritten it to connect it to that answer.

The discussion below treats this structure as a linkage with two fixed vertices and three fixed angles, allowing for variable lengths.

Suppose the vertices are labeled $ABCD$ clockwise from the lower left. Suppose the diagonal $BD$ is fixed (the linked answer fixes edge $AB$). The prescribed angles at those points determine the rays $R$ and $S$ in the directions from $B$ toward $A$ and from $D$ toward $C$.

Now think of $A$ as a variable point on $R$. The known angle at $A$ determines the rat from $A$ toward $C$, which will be the intersection of that ray with $S$. As $A$ moves continuously from near $B$ farther and farther along $R$ angle $ACB$ varies continuously. There will be isolated positions for $A$ that makes that angle correct. I originally thought there was just one such point. Clearly there can be two.

I wonder whether the two extraneous roots of the quartic in the other answer correspond to linkages that somehow use the rays opposite to $R$ and $S$, perhaps leading to nonconvex quadrilaterals.

The more general problem is to think about the shape of the four dimensional set of configurations of this linkage. There will be an open set on which the OP's four angles are local coordinates. What is the singular locus? The boundary?