Let the pentagon have the vertices $A$, $B$, $C$, $D$ and $E$ in this order. We first assume that all sides are of equal length and two angles equal.

If the two equal angles are adjacent, we can place them at $A$ and $B$ without loss of generality. In this case, $EABC$ is an equilateral trapezoid with the common bisector of $AB$ and $EC$ as axis of symmetry. Since $\triangle CDE$ is isosceles with base $CE$, this line is also the axis of symmetry of $\triangle CDE$ and therefore of the entire pentagon $ABCDE$, as required.

If the two equal angles are not adjacent, we can place them at $C$ and $E$. Since triangle $ADE$ and $BDC$ are congruent (SAS) in this case, segments $AD$ and $BD$ are of equal length, and triangle $ABD$ is isosceles with base $AB$. The bisector of $AB$ is therefore an axis of symmetry of $ABD$, and since reflection on this line exchanges $AD$ and $BD$, such a reflection also exchanges the congruent isosceles triangles $ADE$ and $BDC$. It follows that the bisector of $AB$ is the axis of symmetry for the entire pentagon $ABCDE$ as required.

We now assume that all angles are equal and two sides are of equal length.

If the two sides are adjacent, we can place them at $AB$ and $AE$. The triangle $ABE$ is then isosceles with base $BE$, and the bisector of $BE$ is the axis of symmetry of $ABE$. Furthermore, it immediately follows that $BCDE$ is an isosceles trapezoid, since $\angle BCD = \angle EDC$ and $\angle CBE = \angle CBA - \angle EBA = \angle DEA - \angle BEA = \angle DEB$ hold, and this bisector is therefore also the axis of symmetry of this trapezoid, and therefore of the entire pentagon $ABCDE$.
Finally, if the two sides are not adjacent, we can place them at $BC$ and $DE$. Again, $BCDE$ is an isosceles trapezoid and therefore $ABE$ isosceles, with common axis of symmetry in the bisector of $BE$, and this case is seen as analogous to the previous case.

Summing up, we see that every very special pentagon does indeed have an axis of symmetry, as claimed. $\square$