Identity function that returns $1$ for the input $0$.

How about $$f(x) = x + 0^{|x|}$$

As @JohnHughes points out, normal algebra can't help you. If you're willing to use the Kronecker delta function then $$ f(x) = x + \delta(x,0) $$ does what you want.

This can be achieved merely with the arctan and tangent function. The functions $z_1, z_2, z_3$, return $1$ when $x = 0$, and $0$ when $x \neq 0$. Notice this is different than your $id$ function, but will be used to construct it.

$$ z_1(x) = 2^{\lceil |x|\rceil} \text{ mod } 2 $$

$$ z_2(x) = 1 - |\text{sign}(x)| $$

$$ z_3(x) = 1 - \bigg\lceil\frac{|x|}{|x| + 1}\bigg\rceil $$

The absolute values can also be replaced with squares, i.e.

$$ z_3(x) = 1 - \bigg\lceil\frac{x^2}{x^2 + 1}\bigg\rceil $$

Source: I spent a lot of time trying to write conditional functions without conditionals.

Note: $$\lfloor x \rfloor = (x - 0.5) - \frac{\arctan(\tan(\pi(x - 0.5)))}{\pi}$$

Then we can write $$\lceil x \rceil = -\lfloor -x \rfloor$$

So, we can write $z_3$ merely from arctan and tan: $$ z_3(x) = 1 + \bigg\lfloor - \frac{x^2}{x^2 + 1}\bigg\rfloor = 1 - \frac{x^2}{x^2 + 1} - 0.5 - \frac{\arctan(\tan(\pi(- \frac{x^2}{x^2 + 1} - 0.5)))}{\pi}$$

Then,

$$id(x) = z_3(x) + x\cdot(1 - z_3(x))$$

Not pretty, but is entirely in terms of elementary functions.