Equivalence of different ways of geometrical multiplication

1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
2. Construct the perpendicular at $O$.
3. Construct the semicircle on the diameter $A'B$.
4. Find $H$ at the intersection of the semicircle and the perpendicular. $(OH)^2 = OA'\cdot OB = OA\cdot OB$.
5. Draw line $1H$ and construct a perpendicular to it through $H$.
6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1\cdot OK,$ hence $OK = OA\cdot OB.$

The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).

1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
2. Select on $g$ an arbitrary point $P$ other than the origin.
3. Draw a line through $1$ and $P$.
4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
5. Draw a line through $P$ and $B$.
6. Draw a parallel to that line through $Q$. The intersection with the number line is then $A\times B$.

Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).

With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.

If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.

Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $\frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.