Why must be the additive and multiplicative identities in a field be different?

It is possible to define a field with just one element, which has to be the additive and multiplicative identity at the same time. Most definitions exclude this from being a field. If you have at least two elements in your field and try to make the identities the same you fail. Call the common identity $0$ and the other element $a$. But then $$0 \cdot a=a\\(0-0)\cdot a=a\\0\cdot a - 0 \cdot a=a\\a-a=a\\0=a$$

Let $x$ be any element of your field. Then $1=0$ implies $$ x = 1x = 0x =0. $$ Hence your field has only one element. So, you can allow it, but then your field is necessarily trivial. So, if you want a field with more than one element, you have to have $1 \neq 0$.

If $\Bbb F$ is any field for which

$\mathbf 1_{\Bbb F} = \mathbf 0_{\Bbb F}, \tag 1$

we have, for any $\theta \in \Bbb F$,

$\theta = \theta \mathbf 1_{\Bbb F} = \theta \mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F}, \tag 2$

so $\Bbb F$ has only one element, $\mathbf 0_{\Bbb F}$; if we want to avoid the trivial case

$\Bbb F = \{ \mathbf 0_{\Bbb F} \}, \tag 3$

we must assume that

$\mathbf 1_{\Bbb F} \ne \mathbf 0_{\Bbb F}. \tag 4$

Note Added in Edit, Thursday 16 November 2017 9:52 AM PST: In "the spirit" of Ross Millikan's comment, we have:

$\mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F} + \mathbf 0_{\Bbb F}, \tag 5$

since $\mathbf 0_{\Bbb F}$ is the additive identity. Then

$\theta \mathbf 0_{\Bbb F} = \theta(\mathbf 0_{\Bbb F} + \mathbf 0_{\Bbb F}) = \theta \mathbf 0_{\Bbb F} + \theta \mathbf 0_{\Bbb F}, \tag 6$

by the distributive law, whence

$\theta \mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F}. \tag 7$

End of Note.