We proceed in three steps.

Step 1: We first give a recipe for constructing hexagons of this type. Let $ACE$ be a triangle, with all angles less than $2\pi/3$. Let $D$ be the reflection of $A$ across $CE$; let $F$ be the reflection of $C$ across $EA$; let $B$ be the reflection of $E$ across $AC$. Then, we have $\angle BAF = \angle BAC + \angle CAE + \angle EAF = 3\angle CAE = 3\angle CDE$ and the other analogous angle equalities. Also, we have $AB = AE = DE$ and the other analogous side equalities. Thus, the hexagon satisfies the equations in the problem statement. The diagonals $AD$, $BE$, $CF$ are simply the altitudes of the triangle $ACE$, so they are concurrent at the orthocenter.

Step 2: For a hexagon meeting the conditions of the problem statement, let $\beta$, $\delta$, and $\phi$ be the measures of angles $B$, $D$, and $F$. We claim that these are sufficient to determine $ABCDEF$ up to scaling.
Let $x = AB = DE$, $y = BC = EF$, $z = CD = FA$. Our goal is to show that these lengths are determined up to scale by the given angles. Let $a, b, c, d, e$, and $f$ be unit vectors in the directions of the edges from $A$ to $B$, $B$ to $C$, $C$ to $D$, $D$ to $E$, $E$ to $F$, and $F$ to $A$, respectively. Then, we have the vector identity
$$x(a + d) + y(b + e) + z(c + f) = 0. \tag{1}$$
Notice now that
$$4(\beta + \delta + \phi) = \angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 4\pi,$$
so $\beta + \delta + \phi = \pi$. Also, the fact that opposite sides are not parallel implies that $\pi + 2\beta = \angle D + \angle E + \angle F \neq 2\pi$, so $\beta \neq \pi/2$; likewise $\delta, \phi \neq \pi/2$.
Assuming without loss of generality that the vertices of $ABCDEF$ are labeled in counterclockwise order and making liberal use of the identity $\beta+\delta+\phi = \pi$, we may compute the respective orientations of vectors $b$, $c$, $d$, $e$, and $f$, measured counterclockwise relative to $a$ modulo $2\pi$, to be:
$$\begin{align*} b: & \quad \pi - \beta \\ c: & \quad -\beta - 3\phi \\ d: & \quad -2\phi \\ e: & \quad \pi + 2\delta - \beta \\ f: & \quad 2\delta - \phi - \beta \end{align*}$$
Now, whenever two unit vectors point in directions $\theta$ and $\psi$ which do not differ by $\pi$, their sum is a nonzero vector either pointing in direction $(\theta + \psi)/2$ or direction $(\theta + \psi)/2 + \pi$. It follows that the vectors $a+d, b+e$, and $c+f$ are all nonzero and point in the following directions modulo $\pi$:
$$\begin{align*} a+d: & -\phi \\ b+e: & \delta - \beta \\ c+f: & \delta - 2\phi - \beta. \end{align*}$$
Recalling that $\beta + \delta + \phi = \pi$ and $\beta, \delta, \phi \neq \pi/2$, we see that none of the pairwise differences between $a+d, b+e$, and $c+f$ are multiples of $\pi$. Thus, $a+d, b+e$, and $c+f$ are nonzero vectors, no two of which are collinear. Consequently, condition (1) determines the coefficients $x, y, z$ uniquely up to scale, as required.

Step 3: It suffices to show that the only hexagons meeting the conditions of the problem statement are the ones constructed in Step 1. Construct a hexagon $A_1B_1C_1D_1E_1F_1$ by taking $A_1C_1E_1$ to be a triangle with angles $\beta, \delta$, and $\phi$ and reflecting each vertex across the opposite site as in Step 1. Then, $A_1B_1C_1D_1E_1F_1$ satisfies the conditions of the problem statement and has $\angle B_1 = \beta, \angle D_1 = \delta$, and $\angle F_1 = \phi$, so by Step 2 it is similar to $ABCDEF$, completing the proof.