There is no normal form besides that any subterm using $+,\cdot,0,1$—treating its subterms whose topmost functions are $\oplus,\otimes,\bar0,\bar1$ as black boxes—can be written as a polynomial (i.e., a possibly empty commutative sum of possibly empty commutative products), and vice versa. As the question puts it, “we shall have only the full term algebra of all mixed terms in this language”. This holds even if each structure is individually isomorphic to the complex field, as requested in the question. The way to show this is to look at the term functions in the following structure $(F,+,\times,0,1,\oplus,\otimes,\bar0,\bar1)$.
Let $(F_0,+,\cdot,0,1)$ be a copy of $\mathbb C$. Treating the elements of $F_0$ as formal variables, let $(G_0,\oplus,\otimes,\bar0,\bar1)$ be the algebraic closure of the rational function field $\mathbb Q(F_0)$. Treating the elements of $G_0\let\bez\smallsetminus\bez F_0$ as formal variables, let $(F_1,+,\cdot,0,1)$ be an algebraic closure of the rational function field $F_0(G_0\bez F_0)$. Treating the elements of $F_1\bez G_0$ as formal variables, let $(G_1,\oplus,\otimes,\bar0,\bar1)$ be an algebraic closure of $G_0(F_1\bez G_0)$. Etc. Then let $F=\bigcup_{n\in\omega}F_n=\bigcup_{n\in\omega}G_n$ with the respective structure. (For good measure, we can also define a set $G_{-1}$ as a fixed transcendence basis of $F_0$ over $\mathbb Q$.)
Let me define the normal forms more formally. Let $V$ denote the set of variables. Fix a linear order $<$ on the set of all terms over $V$.
If $X$ is a finite set of terms, define $\sum X$ as follows: write $X=\{x_i:i<n\}$, where $n=|X|$ and $x_0<\dots<x_{n-1}$; put $\sum X=0$ if $n=0$, and $\sum X=((\cdots(x_0+x_1)+\cdots)+x_{n-1})$ if $n\ge1$. Similarly, define $\prod X$, $\oplus X$, and $\otimes X$.
Let $T_+$, $T_\cdot$, $T_\oplus$, and $T_\otimes$ be the smallest sets of terms over $V$ satisfying the following inductive conditions:
if $X\subseteq V\cup T_\oplus\cup T_\otimes$ is finite and $|X|\ne1$, then $\prod X\in T_\cdot$;
if $X\subseteq V\cup T_\cdot\cup T_\oplus\cup T_\otimes$ is finite and $|X|\ne1$, then $\sum X\in T_+$;
if $X\subseteq V\cup T_+\cup T_\cdot$ is finite and $|X|\ne1$, then $\otimes X\in T_\otimes$;
if $X\subseteq V\cup T_\otimes\cup T_+\cup T_\cdot$ is finite and $|X|\ne1$, then $\oplus X\in T_\oplus$.
The set of normal forms is defined as $N=V\cup T_+\cup T_\cdot\cup T_\oplus\cup T_\otimes$.
Proposition: Every term is equal to exactly one term in normal form.
Proof sketch:
Existence: by induction on the complexity of the term. A term whose main function is $+,\cdot,0,1$ can be written as $p(t_0,\dots,t_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients (other than a single variable), and $t_i$ are either variables or terms whose main functions are $\oplus,\otimes,\bar0,\bar1$. We can write each $t_i$ as a normal form in $V\cup T_\oplus\cup T_\otimes$ by the induction hypothesis, and express $p$ as an appropriately bracketed sum of products in the right order to make it a normal form.
Uniqueness: let $v(t)$ denote a fixed evaluation of terms $t$ in $F$ above, where variables are evaluated by distinct elements of $G_{-1}$. We want to show that $v(t)\ne v(s)$ if $t,s\in N$ are distinct.
Again, a term in $T_+\cup T_\cdot$ is of the form $p(t_0,\dots,t_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients other than a single variable, and $t_i$ are subterms of $t$ from $V\cup T_\oplus\cup T_\otimes$. Using this, and the fact that elements of $\bigcup_n(G_n\smallsetminus F_n)$ are algebraically independent over $\mathbb Q$ in $(F,+,\cdot,0,1)$, as well as the dual statements swapping the role of the two sets of operations, we can prove by induction on the complexity of $t\in N$ that $v(t)\in\bigcup_n(F_n\smallsetminus G_{n-1})$ if $t\in T_+\cup T_\cdot$, $v(t)\in\bigcup_n(G_n\smallsetminus F_n)$ if $t\in T_\oplus\cup T_\otimes$, and $v(s)\ne v(t)$ for all $s\ne t$ in $N$ whose complexity does not exceed that of $t$.
Best Answer
If the operations have units ($a * 1 = a$ etc), then that is simply called a commutative algebra: https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument Indeed in that case, $a*b = a \circ b = b * a = b \circ a$.
Otherwise, I don't know if there is an established name. This is an algebra over the Boardman-Vogt tensor product of operads $\mathrm{Ass} \otimes_{BV} \mathrm{Ass}$, where $\mathrm{Ass}$ is the operad encoding associative algebras. This is not an uninteresting structure, and it exhibits nontrivial hidden commutativity properties. Here is a summary taken from Boardman--Vogt tensor products of absolutely free operads (Bremner-Dotsenko)
The quoted results are from:
If I had to choose a name, I would call them "double algebras", but I do not believe this is standard. "Double semigroup" certainly is standard terminology, but there is not requirement of linearity.