Suppose $x \oplus y$ is of the form $h\big(g(x) + g(y)\big)$ with $g\big(h(x)\big) = x$, which immediately guarantees commutativity and associativity: $x \oplus y = h\big(g(x) + g(y)\big) = h\big(g(y) + g(x)\big) = y \oplus x,$ $(x\oplus y)\oplus z = h\big(g(h\big(g(x) + g(y)\big))+g(z)\big) = h\big(g(x) + g(y) + g(z)\big) = h\big(g(x) + g(h\big(g(y) + g(z)\big))\big) = x\oplus(y\oplus z).$ Now let $g(x)$ be the complex number with the same modulus as $x$ and twice its argument. That is, $g(0) = 0$, and $g(x) = \lvert x\rvert e^{2i\arg x}$ when $x \ne 0$. Then we can set $h(0) = 0$ and $h(x) = \lvert x\rvert e^{i(\arg x)/2}$ otherwise. Assuming $\arg$ has range $(-\pi,\pi]$, this has a discontinuity along the negative real line, and we have $g\big(h(x)\big) = x$, but $h\big(g(x)\big) = x \operatorname{sgn} \operatorname{Re} x$. Now observe that $g(\pm ix) = e^{\pm i\pi}\lvert x\rvert e^{2i\arg x} = -g(x),$ and for real $a$, $\arg ax$ is either $\arg x$ or $\arg x \pm \pi$ depending on the sign of $a$, but in either case $g(ax) = \lvert a\rvert\lvert x\rvert e^{2i\arg x}e^{\pm2i\pi} = \lvert a\rvert g(x).$
Now we can verify the third and (original) fourth conditions: $x\oplus(\pm ix) = h\big(g(x) + g(\pm ix)\big) = h\big(g(x) - g(x)\big) = h(0) = 0.$ $(ax)\oplus(bx) = h\big(g(ax)+g(bx)\big) = h\big(\lvert a\rvert g(x) + \lvert b\rvert g(x)\big) = h\big((\lvert a\rvert + \lvert b\rvert)g(x)\big) = h\big(g((\lvert a\rvert+\lvert b\rvert)x)\big) = h\big(g((\lvert a\rvert+\lvert b\rvert)x) + g(0)\big) = (\lvert a\rvert+\lvert b\rvert)x \oplus 0.$
(I don't understand why the original fourth condition is troublesome or implies that $x \oplus 0 = 0$. Perhaps there is an error in this answer.)