TITLE: Commutative binary operations on $\Bbb C$ that distribute over both multiplication and addition
QUESTION [8 upvotes]: Does there exist a non-trivial commutative binary operation on $\Bbb C$ that distributes over both multiplication and addition?
In other words, if our operation is denoted by $\odot$, then I want the following to hold:

$a \odot (b \cdot c) = a \odot b \cdot a \odot c$
$a \odot (b + c) = a \odot b + a \odot c$
$a \odot b = b \odot a$

All of the things I can find so far distribute over either multiplication or addition, but not both. Alternatively, is there a proof that no such operation can exist?
I wasn't sure if this question was too elementary for MO, so I'm trying here first. This question is obliquely related to the following other questions I've asked on MO and here:

A particularly “natural” algebraic structure with three commutative, pairwise-distributive operations
Translations AND dilations of infinite series
Discrete “difference” equations that involve changes in both shift and scale

REPLY [5 votes]: To add to Eric's answer above, it doesn't even require the existence of a multiplicative identity to shoot this down. See below:
$2[a\odot(b \cdot c)] \\
= a\odot(b \cdot c) + a\odot(b \cdot c) \\
= (a+a) \odot (b \cdot c) \\
= [(a+a) \odot b] \cdot [(a+a) \odot c] \\
= [(a \odot b) + (a \odot b)] \cdot [(a \odot c) + (a \odot c)] \\
= (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) \\
= a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) \\
= 4[a \odot (b \cdot c)]$