I argue that the expression for infimum is more complicated and cannot be given by a single formula unless certain inequalities involving $a,b,c,d$ are assumed.
Let us work in the setting that OP considered for infimum: $a<b<c<d$. Denote the angle between edges of lengths $a,b$ by $\alpha$ and the one between edges of lengths $c,d$ by $\beta$. They are related only by the following constraint (which comes from considering the diagonal length $e$ in the picture):
$$a^2+b^2-2ab\cos\alpha=c^2+d^2-2cd\cos\beta.$$
Thus the goal is to obtain the optima of the area function $f(\alpha,\beta)=\frac{1}{2}ab\sin\alpha+\frac{1}{2}cd\sin\beta$ subject to the constraint $g(\alpha,\beta)=(a^2+b^2-2ab\cos\alpha)-(c^2+d^2-2cd\cos\beta)=0$. Using Lagrange multipliers, let us consider the points where $\nabla f=\left(\frac{1}{2}ab\cos\alpha,\frac{1}{2}cd\cos\beta\right)$ is a multiple of $\nabla g=(2ab\sin\alpha,-2cd\sin\beta)$. This amounts to $\tan\alpha=-\tan\beta$, i.e. $\alpha+\beta=\pi$. This is the case of maximum area mentioned in other answers. Therefore, to derive the infimum of $f(\alpha,\beta)$ given $g(\alpha,\beta)=0$, we should consider boundary values for $\beta$. The assumptions $d>a$ and $c>b$ imply $\beta<\alpha$. Also $\beta+\alpha\leq 2\pi$. (The sum of all four angles is $2\pi$ even if the quadrilateral is concave.) Thus $\beta$ belongs to $[0,\pi]$. The boundary values for $\beta$ are thus uniquely determined by those of $\cos\beta$. Based on constraint, the latter yield $\cos\alpha$ and hence two opposite choices for $\sin\alpha$. Substituting the negative one in the area formula yields a candidate for the minimum. In view of this discussion, let's analyze the extreme values that $\cos\beta$ takes.
Solving the equation above for $\cos\beta$, the fraction $\frac{c^2+d^2-a^2-b^2+2ab\cos\alpha}{2cd}$ must be between $-1$ and $1$. Conversely, any such fraction in $[-1,1]$ determines $\beta$ uniquely. The range for $\cos\beta=\frac{c^2+d^2-a^2-b^2+2ab\cos\alpha}{2cd}$ is
$$\left[\max\left(\frac{c^2+d^2-a^2-b^2-2ab}{2cd},-1\right), \min\left(\frac{c^2+d^2-a^2-b^2+2ab}{2cd},1\right)\right].$$
Notice that: $\frac{c^2+d^2-a^2-b^2-2ab}{2cd}\geq -1\Leftrightarrow d\geq a+b+c$, $\frac{c^2+d^2-a^2-b^2-2ab}{2cd}\geq 1\Leftrightarrow a+d\geq b+c$. So to simplify further, three different regimes should be considered:
- $d\geq a+b+c$ (in which case the quadrilateral is concave unlike the picture above),
- $b+c-a<d\leq a+b+c$ (this pertains to the limit case that OP mentions: $\beta$ attains all values in $[0,\pi]$, and $\beta=0$ yields the minimum possible area as the area of a triangle with sides $a,b,d-c$);
- $d\leq b+c-a$.
Conclusion) In general, further assumptions are needed to obtain a formula for the infimum of area, because there are three different regimes. The second one is easier to handle. For the first one $(d\leq a+b+c)$, the minimum area is achieved when $\beta=0$ or $\beta=\cos^{-1}\left(\frac{c^2+d^2-a^2-b^2-2ab}{2cd}\right)$. In the third regime $(d\leq b+c-a)$ the minimum area is achieved when $\beta=\cos^{-1}\left(\frac{c^2+d^2-a^2-b^2+2ab}{2cd}\right)$ or $\beta=\cos^{-1}\left(\frac{c^2+d^2-a^2-b^2-2ab}{2cd}\right)$.
Note) I have assumed that the quadrilateral does not cross itself. Indeed, the Bretschneider's formula does not hold in general if it does, and even the expression for the maximum area becomes complicated in that situation. But I have not assumed anything about convexity.