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I think you must have misunderstood how Lawvere theories are supposed to work. Let me try to motivate them from an algebraist point of view, and answer your question in passing.

An algebraic structure such as a group or a ring is usually described in terms of operations and equations they satisfy. But why do we choose certain operations and certain equations? There are many ways of axiomatizing any given algebraic structure. For example, in a ring we could take the unary operation negation $-x$ as basic, or the binary operation subtraction $x - y$ as basic. And why not both? Why is multiplication taken as basic in the theory of groups, rather than division?

Can we describe an algebraic structure in a canonical way, so that no preference is made about which operations and equations count as "basic"? If we are not allowed to prefer any particular choice of operations and axioms, then we must favor them all equally! So, a "canonical" description should include all operations and all equations.

Let us consider the theory of groups to see how this works. We customarily start with three basic operations: a nullary operation (constant) unit $1$, a unary operation inverse ${}^{-1}$, and a binary operation multiplication $\cdot$. There are five axioms:

• $x \cdot (y \cdot z) = (x \cdot y) \cdot z$
• $x \cdot 1 = x$
• $1 \cdot x = x$
• $x \cdot x^{-1} = 1$
• $x^{-1} \cdot x = 1$

To get a canonical theory of groups which does not prefer the above operations and axioms, we should generate all possible operations and equations out of the basic ones, and then take them all as basic. Some generated operations will have names, e.g., division $x/y = x \cdot y^{-1}$, squaring $x^2 = x \cdot x$, etc., but others will not. An example of such an operation might be a ternary operation $p(x_1,x_2,x_3) = ((x_1 \cdot 1) \cdot x_2^{-1}) \cdot (x_3 \cdot x_2)$. In fact, for each $n$ there will be infinitely many $n$-ary operations, represented by expressions built from basic operations and variables $x_1, \ldots, x_n$. As new "axioms" we will simply take all equations that follow logically from the above five axioms. If we now "forget" that we started with the three basic operations and five axioms, we will get a somewhat unusual theory with infinitely many operations and infinitely many axioms, which nevertheless still describes what a group is.

This all sounds very messy. Are we supposed to invent notation for infinitely many operations? And what use is there in having so many axioms that they already are closed under logical deduction? If we truly are algebraists at heart, we must free ourselves of the shackles of syntax.

Let us make a category $\mathcal{T}$ out of the messy description of groups we generated above. Remember that we are not trying to construct the category of groups, but rather a category which nicely organizes the infinitely many operations and equations. The idea is simple enough. The morphisms of $\mathcal{T}$ should correspond to the operations of the theory. The equations should correspond to the fact that certain morphisms are equal.

An $n$-ary operation can be thought of as a map $G^n \to G$ where $G$ is the carrier set of a group and $G^n$ is the $n$-fold product of $G$'s. Thus, for each $n$ there should be an object $\mathtt{G}^n$ in $\mathcal{T}$ and the morphisms $\mathtt{G}^n \to \mathtt{G}^1$ should correspond to the $n$-ary operations. We therefore posit that the objects of $\mathcal{T}$ are $$\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$$ where you must not think of $\mathtt{G}^n$ as any kind of set, or an $n$-fold product of anything. We formally write $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$, but we could have as well declared that the objects of $\mathcal{T}$ are the natural numbers and write them simply but confusingly as $0, 1, 2, \ldots$.

The morphisms $\mathtt{G}^n \to \mathtt{G}^1$ are expressions built from $n$ variables $x_1, \ldots, x_n$ and the operations $1$, $\cdot$, and ${}^{-1}$. Two such expressions $p(x_1, \ldots, x_n)$ and $q(x_1, \ldots, x_n)$ are considered to be the same morphism if the theory of groups proves $p(x_1, \ldots, x_n) = q(x_1, \ldots, x_n)$. What about the morphisms $\mathtt{G}^n \to \mathtt{G}^m$? Since we expect that $\mathtt{G}^m$ will in fact end up being the $m$-fold product of $\mathtt{G}^1$'s, a morphism $\mathtt{G}^n \to \mathtt{G}^m$ corresponds uniquely to an $m$-tuple of morphisms $\mathtt{G}^n \to \mathtt{G}^1$. In other words, the morphisms $\mathtt{G}^n \to \mathtt{G}^m$ are $m$-tuples of expressions in variables $x_1, \ldots, x_n$, where again two such $m$-tuples represent the same morphism if the theory of groups proves them equal.

Composition in $\mathcal{T}$ is performed by substitution. For example, the composition of $(x_1^{-1}, x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^2$ and $(x_1 \cdot x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$ is $(x_1^{-1} \cdot x_1^{-1} \cdot (x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^1$. The identity morphism from $\mathtt{G}^n \to \mathtt{G}^n$ is the $n$-tuple $(x_1, x_2, \ldots, x_n)$.

You might think that a better $\mathcal{T}$ would contain just three objects $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2$ and morphisms corresponding to the basic operations $1$, ${}^{-1}$ and $\cdot$. But that would not even be a category, and if somehow you managed to make one, you would still have to make sure that $\mathtt{G}^2 = \mathtt{G}^1 \times \mathtt{G}^1$ and that $\mathtt{G}^0$ is the terminal object. It would not really be any prettier.

We can now actually show that $\mathtt{G}^m$ is the $m$-fold product of $\mathtt{G}^{1}$'s. The $k$-th projection $\pi_k : \mathtt{G}^m \to \mathtt{G}^1$ is (represented by) the expression $x_k$. I leave the rest as an exercise. Another exercise is to show that $\mathcal{T}$ has finite products computed as $\mathtt{G}^n \times \mathtt{G}^m = \mathtt{G}^{n + m}$.

We are still doing a lot of syntax disguised as category theory, but that is a necessary step that allows us to see what sort of category $\mathcal{T}$ is. We are ready to define when a category in general is the description of an algebraic theory. I am phrasing the following definition a bit imprecisely without the technical distraction of requiring a "strict identity on objects functor from $\aleph_0^\mathrm{op}$ ...":

Definition [Lawvere]: An algebraic theory is a category with distinct objects $\mathtt{A}^0, \mathtt{A}^1, \mathtt{A}^2, \ldots$ such that $\mathtt{A}^n$ is the $n$-fold product of $\mathtt{A}^1$'s.

It turns out that the models of such a theory/category $\mathcal{T}$ are precisely those functors $F : \mathcal{T} \to \mathsf{Set}$ which preserve finite products. Every such functor is already determined by how it maps $\mathtt{A}^1$ and morphisms $\mathtt{A}^n \to \mathtt{A}^1$, which of course correspond to the $n$-ary operations. In a particular case $F$ might be determined by even fewer pieces of information. For example, if $\mathcal{T}$ is the category which describes the theory of groups, $F$ will be determined already by how it maps the morphisms $1 : \mathtt{G}^0 \to \mathtt{G}^1$, $(x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$, and $(x_1^{-1}) : \mathtt{G}^1 \to \mathtt{G}^1$, because these generate all other morphisms (except projections and pairings, but $F$ preserves products).

The whole point of the exercise was to arrive at a non-syntactic notion of "algebraic theory". The next step is to look for examples which really are non-syntactic in nature. Here is one: the category whose objects are Euclidean spaces $\mathbb{R}^n$ and the morphisms are smooth maps. This theory describes what is known as smooth algebras.

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I think you must have misunderstood how Lawvere theories are supposed to work. Let me try to motivate them from an algebraist point of view, and answer your question in passing.

An algebraic structure such as a group or a ring is usually described in terms of operations and equations they satisfy. But why do we choose certain operations and certain equations? There are many ways of axiomatizing any given algebraic structure. For example, in a ring we could take the unary operation negation $-x$ as basic, or the binary operation subtraction $x - y$ as basic. And why not both? Why is multiplication taken as basic in the theory of groups, rather than division?

Can we describe an algebraic structure in a canonical way, so that no preference is made about which operations and equations count as "basic"? If we are not allowed to prefer any particular choice of operations and axioms, then we must favor them all equally! So, a "canonical" description should include all operations and all equations.

Let us consider the theory of groups to see how this works. We customarily start with three basic operations: a nullary operation (constant) unit $1$, a unary operation inverse ${}^{-1}$, and a binary operation multiplication $\cdot$. There are five axioms:

• $x \cdot (y \cdot z) = (x \cdot y) \cdot z$
• $x \cdot 1 = x$
• $1 \cdot x = x$
• $x \cdot x^{-1} = 1$
• $x^{-1} \cdot x = 1$

To get a canonical theory of groups which does not prefer the above operations and axioms, we should generate all possible operations and equations out of the basic ones, and then take them all as basic. Some generated operations will have names, e.g., division $x/y = x \cdot y^{-1}$, squaring $x^2 = x \cdot x$, etc., but others will not. An example of such an operation might be a ternary operation $p(x_1,x_2,x_3) = ((x_1 \cdot 1) \cdot x_2^{-1}) \cdot (x_3 \cdot x_2)$. In fact, for each $n$ there will be infinitely many $n$-ary operations, represented by expressions built from basic operations and variables $x_1, \ldots, x_n$. As new "axioms" we will simply take all equations that follow logically from the above five axioms. If we now "forget" that we started with the three basic operations and five axioms, we will get a somewhat unusual theory with infinitely many operations and infinitely many axioms, which nevertheless still describes what a group is.

This all sounds very messy. Are we supposed to invent notation for infinitely many operations? And what use is there in having so many axioms that they already are closed under logical deduction? If we truly are algebraists at heart, we must free ourselves of the shackles of syntax.

Let us make a category $\mathcal{T}$ out of the messy description of groups we generated above. Remember that we are not trying to construct the category of groups, but rather a category which nicely organizes the infinitely many operations and equations. The idea is simple enough. The morphisms of $\mathcal{T}$ should correspond to the operations of the theory. The equations should correspond to the fact that certain morphisms are equal.

An $n$-ary operation can be thought of as a map $G^n \to G$ where $G$ is the carrier set of a group and $G^n$ is the $n$-fold product of $G$'s. Thus, for each $n$ there should be an object $\mathtt{G}^n$ in $\mathcal{T}$ and the morphisms $\mathtt{G}^n \to \mathtt{G}^1$ should correspond to the $n$-ary operations. We therefore posit that the objects of $\mathcal{T}$ are $$\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$$ where you must not think of $\mathtt{G}^n$ as any kind of set, or an $n$-fold product of anything. We formally write $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$, but we could have as well declared that the objects of $\mathcal{T}$ are the natural numbers and write them simply but confusingly as $0, 1, 2, \ldots$.

The morphisms $\mathtt{G}^n \to \mathtt{G}^1$ are expressions built from $n$ variables $x_1, \ldots, x_n$ and the operations $1$, $\cdot$, and ${}^{-1}$. Two such expressions $p(x_1, \ldots, x_n)$ and $q(x_1, \ldots, x_n)$ are considered to be the same morphism if the theory of groups proves $p(x_1, \ldots, x_n) = q(x_1, \ldots, x_n)$. What about the morphisms $\mathtt{G}^n \to \mathtt{G}^m$? Since we expect that $\mathtt{G}^m$ will in fact end up being the $m$-fold product of $\mathtt{G}^1$'s, a morphism $\mathtt{G}^n \to \mathtt{G}^m$ corresponds uniquely to an $m$-tuple of morphisms $\mathtt{G}^n \to \mathtt{G}^1$. In other words, the morphisms $\mathtt{G}^n \to \mathtt{G}^m$ are $m$-tuples of expressions in variables $x_1, \ldots, x_n$, where again two such $m$-tuples represent the same morphism if the theory of groups proves them equal.

Composition in $\mathcal{T}$ is performed by substitution. For example, the composition of $(x_1^{-1}, x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^2$ and $(x_1 \cdot x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$ is $(x_1^{-1} \cdot x_1^{-1} \cdot (x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^1$. The identity morphism from $\mathtt{G}^n \to \mathtt{G}^n$ is the $n$-tuple $(x_1, x_2, \ldots, x_n)$.

You might think that a better $\mathcal{T}$ would contain just three objects $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2$ and morphisms corresponding to the basic operations $1$, ${}^{-1}$ and $\cdot$. But that would not even be a category, and if somehow you managed to make one, you would still have to make sure that $\mathtt{G}^2 = \mathtt{G}^1 \times \mathtt{G}^1$ and that $\mathtt{G}^0$ is the terminal object. It would not really be any prettier.

We can now actually show that $\mathtt{G}^m$ is the $m$-fold product of $\mathtt{G}^{1}$'s. The $k$-th projection $\pi_k : \mathtt{G}^m \to \mathtt{G}^1$ is (represented by) the expression $x_k$. I leave the rest as an exercise. Another exercise is to show that $\mathcal{T}$ has finite products computed as $\mathtt{G}^n \times \mathtt{G}^m = \mathtt{G}^{n + m}$.

We are still doing a lot of syntax disguised as category theory, but that is a necessary step that allows us to see what sort of category $\mathcal{T}$ is. We are ready to define when a category in general is the description of an algebraic theory. I am phrasing the following definition a bit imprecisely without the technical distraction of requiring a "strict identity on objects functor from $\aleph_0^\mathrm{op}$ ...":

Definition [Lawvere]: An algebraic theory is a category with distinct objects $\mathtt{A}^0, \mathtt{A}^1, \mathtt{A}^2, \ldots$ such that $\mathtt{A}^n$ is the $n$-fold product of $\mathtt{A}^1$'s.

It turns out that the models of such a theory/category $\mathcal{T}$ are precisely those functors $F : \mathcal{T} \to \mathsf{Set}$ which preserve finite products. Every such functor is already determined by how it maps morphisms $\mathtt{A}^n \to \mathtt{A}^1$, which of course correspond to the $n$-ary operations. In a particular case $F$ might be determined by even fewer pieces of information. For example, if $\mathcal{T}$ is the category which describes the theory of groups, $F$ will be determined already by how it maps the morphisms $1 : \mathtt{G}^0 \to \mathtt{G}^1$, $(x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$, and $(x_1^{-1}) : \mathtt{G}^1 \to \mathtt{G}^1$, because these generate all other morphisms (except projections and pairings, but $F$ preserves products).

The whole point of the exercise was to arrive at a non-syntactic notion of "algebraic theory". The next step is to look for examples which really are non-syntactic in nature. Here is one: the category whose objects are Euclidean spaces $\mathbb{R}^n$ and the morphisms are smooth maps. This theory describes what is known as smooth algebras.

You might think that a better $\mathcal{T}$ would contain just three objects $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2$ and morphisms corresponding to the basic operations $1$, ${}^{-1}$ and $\cdot$. But that would not even be a category, and if somehow you managed to make one, you would still have to make sure that $\mathtt{G}^2 = \mathtt{G}^1 \times \mathtt{G}^1$ and that $\mathtt{G}^0$ is the terminal object. It would not really be any prettier.
Definition [Lawvere]: An algebraic theory is a category $\mathcal{T}$ with distinct objects $\mathtt{A}^0, \mathtt{A}^1, \mathtt{A}^2, \ldots$ such that $\mathtt{A}^n$ is the $n$-fold product of $\mathtt{A}^1$'s.
It turns out that the models of such a theory theory/category $\mathcal{T}$ are precisely those functors $F : \mathcal{T} \to \mathsf{Set}$ which preserve finite products. Every such functor is already determined by how it maps morphisms $\mathtt{A}^n \to \mathtt{A}^1$, which of course correspond to the $n$-ary operations. In a particular case $F$ might be determined by even fewer pieces of information. For example, if $\mathcal{T}$ is the category which describes the theory of groups, $F$ will be determined already by how it maps the morphisms $1 : \mathtt{G}^0 \to \mathtt{G}^1$, $(x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$, and $(x_1^{-1}) : \mathtt{G}^1 \to \mathtt{G}^1$, because these generate all other morphisms (except projections and pairings, but $F$ preserves products).