I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.

Choose a base in $\mathbb R^n$ so that all nonzero elements of $\mathbb Z^n$ have nonzero yet algebraic coordinates in this base. For thse coordinates, define a coordinatewise multiplication $*$. Finally, define $x\circ y=(ex)*y$. The operation $\circ$ is what you need since no nonzero vector in $\mathbb Z^n$ is a product of two other such vectors.

Example. If $n=1$, then the first step is unnecessary: one may just define $x\circ y=exy$.

For $n>1$, the first step is needed to avoid large annihilators of nonzero elements of $\mathbb Z^n$. Say, for $n=2$ we introduce a base $(1,\sqrt2)$, $(\sqrt2,1)$. The new coordinates of a vector $(x,y)\in\mathbb R^2$ are $(x',y')=(y\sqrt2-x,x\sqrt2-y)$. Now, given $(x_1,y_1),(x_2,y_2)\in\mathbb R^2$, we take their new coordinates $(x_1',y_1')$ and $(x_2',y_2')$ and multiply them coordinatewise to get $(x_1'x_2',y_1'y_2')$ which are the new coordinates of $(x_1,y_1)*(x_2,y_2)$. So its old coordinates are $(x_1'x_2'+y_1'y_2'\sqrt2,x_1'x_2'\sqrt2+y_1'y_2')$. Multiplying them by $e$ we get $(x_1,y_1)\circ(x_2,y_2)$.

In the new coordinates, this operation is just a componentwise multiplication (conjugated by a multiplication by $e$), so it is associative and distributive. THe change of coordinates is performed to achieve that all coordinates of a nonzero vector from $\mathbb Z^n$ are nonzero.

I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.

Let me describe the construction more formally. Let $\circ$ be a componentwise multiplicaion on $\mathbb R^n$, i.e., $(x_1,\dots,x_n)\circ(y_1,\dots,y_n)=(x_1y_1,\dots,x_ny_n)$.

Choose a no-degenerate matrix $S$ such that each its column consists of $n$ algebraic numbers linearly independent over $\mathbb Q$. This ensures that for every nonzero $x\in\mathbb Z^n$ all the components of $xS$ are nonzero.

Now define $$ x*y=e\bigl((xS)\circ(yS)\bigr)S^{-1}, $$ where $e$ is the base of natural logarithms. We claim that $*$ is a desired operation. One may easily see that the algebras $(\mathbb R,+,*)$ and $(\mathbb R,+,\circ)$ are isomorphic, the isomorphism is provided by $x\mapsto exS$. Thus $*$ is associative and distributive over addition.

Next, for every nonzero $x,y\in\mathbb Z^n$, the product $(xS)\circ(yS)$ consists of nonzero algebraic numbers, so $\bigl((xS)\circ(yS)\bigr)S^{-1}$ is a nonzero vector with algebraic components. Thus $x*y$ contains at least one transcendent component and thus does not lie in $\mathbb Z^n$. Therefore, the conditions 1) and 2) are trivially satisfied (for nonzero $z$).

I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.

Choose a base in $\mathbb R^n$ so that all nonzero elements of $\mathbb Z^n$ have nonzero yet algebraic coordinates. (It suffices to choose a base consisting of vectors whose coordinates are $\mathbb Q$-independent algebraic numbers). In a new base, refine a coordinatewise multiplication $*$. Finally, define $x\circ y=(ex)*y$. The operation $\circ$ is what you need since no nonzero vector in $\mathbb Z^n$ is a product of two other such vectors.

I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.

Choose a base in $\mathbb R^n$ so that all nonzero elements of $\mathbb Z^n$ have nonzero yet algebraic coordinates in this base. For thse coordinates, define a coordinatewise multiplication $*$. Finally, define $x\circ y=(ex)*y$. The operation $\circ$ is what you need since no nonzero vector in $\mathbb Z^n$ is a product of two other such vectors.

Example. If $n=1$, then the first step is unnecessary: one may just define $x\circ y=exy$.

For $n>1$, the first step is needed to avoid large annihilators of nonzero elements of $\mathbb Z^n$. Say, for $n=2$ we introduce a base $(1,\sqrt2)$, $(\sqrt2,1)$. The new coordinates of a vector $(x,y)\in\mathbb R^2$ are $(x',y')=(y\sqrt2-x,x\sqrt2-y)$. Now, given $(x_1,y_1),(x_2,y_2)\in\mathbb R^2$, we take their new coordinates $(x_1',y_1')$ and $(x_2',y_2')$ and multiply them coordinatewise to get $(x_1'x_2',y_1'y_2')$ which are the new coordinates of $(x_1,y_1)*(x_2,y_2)$. So its old coordinates are $(x_1'x_2'+y_1'y_2'\sqrt2,x_1'x_2'\sqrt2+y_1'y_2')$. Multiplying them by $e$ we get $(x_1,y_1)\circ(x_2,y_2)$.

In the new coordinates, this operation is just a componentwise multiplication (conjugated by a multiplication by $e$), so it is associative and distributive. THe change of coordinates is performed to achieve that all coordinates of a nonzero vector from $\mathbb Z^n$ are nonzero.