Examine my argument that $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$ Setting
Let $\mathcal{L} = \{+,0\}$. I want to show that $$\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$$. Note $\equiv$ means elementary equivalence in this question.

Updated Problem

My issue is that given the four ring axioms for addition and identity:

$$ \forall x \forall y \forall z ~  (x + y) + z = x + (y + z)\\
 \forall x        x + 0 = 0 + x = x\\
 \forall x \exists y     x + y = y + x = 0\\
 \forall x \forall y     x + y = y + x$$

it seems like $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$. As someone commented below if I can use multiplication then I may prove $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$, but multiplication is not in $\mathcal{L}$.
Here is Brian Scott's sentence, slightly simplified.

$$\forall x\forall y\exists z ((z+z=x)\lor (z+z=y)\lor (z+z=x+y))$$

This is true in $\mathbb Z$ - it essentially says that given $x,y$, at least one of $x,y,x+y$ is even.

It is not true for $\mathbb Z\oplus\mathbb Z$, for example if $x=(1,0),y=(0,1)$.
I am not quite what the question is here. But I will answer the question

Show that there does not exist a bijective group homomorphism (isomorphism) between $Z$ and $Z \times Z.$

Note that every element of $Z$ can be written as $n.1$ or $-(n.1)$ where $n.$ means add $n$ copies of together and $-$ means take the additive inverse. i.e. $1$ generates the group.

$Z\times Z$ does not have this property for any element of it. If we take $(a,0)$ with $a \neq 0,$ as our
generator we never get non-zero in the second coordinate. If we take $(0,b)$ we never get it the first, If we take $(a,b)$ with both non-zero, we'll never zero in either apart from $(0,0).$ 

So no element generates. 

Since the property of having a group element generate is defined purely in terms of the group actions, it would be preserved by an isomorophism. Hence no isomorphism exists.

(I discuss more of these sorts of arguments in my forthcoming book Proof Patterns. )