Commutative binary operations on $\Bbb C$ that distribute over both multiplication and addition

One such operation is $a\odot b=0$ for all $a,b$. I claim this is the only such operation. Indeed, we have $$a\odot c=a\odot(1\cdot c)=(a\odot 1)\cdot (a\odot c).$$ Taking $c=1$ gives that $a\odot 1$ must be either $0$ or $1$ for each $a$. But if $a\odot 1=1$, then $(a+a)\odot 1=2$, which is impossible. So in fact $a\odot 1=0$ for all $a$, and now the equation above tells us $a\odot c=0$ for all $c$ as well.

This argument uses only the fact that $\odot$ distributes over multiplication on the left and $\odot$ distributes over addition on the right. With slight modification, it applies equally well with $\mathbb{C}$ replaced by any ring in which $2$ is not a zero divisor. Note that in arbitrary rings, there can be other such operations $\odot$. For instance, in a Boolean ring (in which $a\cdot a=a$ for all $a$), $a\odot b=a\cdot b$ is such an operation.

To add to Eric's answer above, it doesn't even require the existence of a multiplicative identity to shoot this down. See below:

$2[a\odot(b \cdot c)] \\ = a\odot(b \cdot c) + a\odot(b \cdot c) \\ = (a+a) \odot (b \cdot c) \\ = [(a+a) \odot b] \cdot [(a+a) \odot c] \\ = [(a \odot b) + (a \odot b)] \cdot [(a \odot c) + (a \odot c)] \\ = (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) \\ = a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) \\ = 4[a \odot (b \cdot c)]$