There exist many algebraic surfaces of constant breadth with no continuous symmetries, even ones with no symmetries at all. To see this, consider the properties of the support parametrization:

** The support parametrization.**
Let $B\subset\mathbb{R}^3$ be a smooth, strongly convex surface, i.e., the outward Gauss map $u:B\to S^2$ is a diffeomorphism. Since it is a diffeomorphism, there is a smooth inverse $b:S^2 \to B$ that is also a diffeomorphism. Since, for $u\in S^2$, the (supporting) tangent plane to $B$ at $b(u)$ is orthogonal to $u$, it can be written in the form $x\cdot u = p(u)$, where $p: S^2\to \mathbb{R}$ is a smooth function on $S^2$. Then, of course, one has $b(u)\cdot u = p(u)$
and one also has $\mathrm{d}b\cdot u = 0$ (since the differential of $b$ at $u\in S^2$
has its image being the subspace parallel to the tangent plane to $B$ at $b(u)$, which, as we have seen, is perpendicular to $u$). The two equations $b(u)\cdot u = p(u)$ and $\mathrm{d}b\cdot u = 0$ then determine $b$ in terms of $p$: One has the vector equation
$$
b = p\,u + \nabla p,
$$
where $\nabla p$ is the gradient vector field of $p$ as a function on $S^2$. The function $p$ is called the *support function* of $B$, and the above parametrization is called the *support parametrization*. (Of course, there is nothing special about $3$ dimensions here, the support parametrization works for strongly convex hypersurfaces in all dimensions.)

** Surfaces of Constant Breadth.**
Now, consider the two supporting tangent planes to $B$ that are perpendicular to $u$: They are the tangent plane at $b(u)$ and the tangent plane at $b(-u)$. They have equations $x\cdot u = p(u)$ and $x\cdot (-u) = p(-u)$ respectively. In particular, the distance between these two planes is $p(u)+p(-u)$. Thus, $B$ has constant breadth $2d$ if and only if its support function $p$ satisfies $p(u)+p(-u) = 2d$. In particular, the function $p_0=p-d$ must be an *odd* function on $S^2$, i.e., $p_0(-u) = -p_0(u)$ for all $u\in S^2$.

Conversely, if $p_0:S^2\to\mathbb{R}$ is an odd function, then $p = p_0+d$ for $d>0$ sufficiently large is then the support function of a strongly convex body of constant breadth $2d$. Alternatively, for all sufficiently small $\epsilon>0$, the function $p = 1 + \epsilon p_0$ is the support function of a strongly convex surface $B_\epsilon$ of constant breadth $2$.

** An algebraic example.**
To construct an algebraic example, it suffices to take $p_0$ to be an odd algebraic function on $S^2$. Taking $p_0$ to be the restriction of a linear function in $\mathbb{R}^3$ to $S^2$ will only give a surface that is a translation of the round $S^2$, so we need to look at something else.

The next simplest choice would be $p_0 = xyz$, where $S^2$ is defined by $x^2+y^2+z^2=1$. For $\epsilon^2<1$, the surface $B_\epsilon=b_\epsilon(S^2)$ is a surface of constant breadth $2$ that is smooth, strongly convex, and algebraic. (When $\epsilon^2=1$, the surface $B_1$ is still strictly convex and algebraic and has breadth $2$, but it is not smooth. When $\epsilon^2>1$ the image $b_\epsilon(S^2)$ is not a convex surface.) When $\epsilon\not=0$, it is easy to show that $B_\epsilon$ has the symmetry of the function $xyz$ but no more symmetry than that, i.e., the symmetry group will have (finite) order $24$ (in fact, it is the symmetry group of the regular tetrahedron.) It is algebraic because it is parametrized by algebraic functions on the algebraic surface $S^2$. It is therefore (a component of) the zero locus of an irreducible polynomial function $f_\epsilon(x,y,z)$ on $\mathbb{R}^3$. A calculation using elimination theory (carried out by MAPLE) shows that $f_\epsilon$ has degree $20$ (when $\epsilon\not=0$) and, for most values of $\epsilon$, has more than 1000 terms.

It is not obvious that the zero locus of $f_\epsilon$ has no other (real) components other than $B_\epsilon$. Thus, I don't know that the interior of $B_\epsilon$ (which makes sense as long as $\epsilon^2\le 1$) can be characterized as the set where $f_\epsilon$ is negative. However, I don't know why you want this; it doesn't seem natural to me. Surely, you only really need it to be algebraic. In any case, one has an explicit parametrization of $B_\epsilon$ using any explicit parametrization of $S^2$. It is easy to draw pictures using graphical software. Here are what the surfaces $B_{1/2}$ and $B_1$ look like:

**Added comment:** David Speyer has shown that $f_\epsilon$ cannot change sign anywhere other than across the image $b_\epsilon(S^2)$, since, as he shows, this is the only two-dimensional component of the real locus of $f_\epsilon=0$.

More generally, taking $p = 1+p_0$, where $p_0 = \lambda_1\lambda_2\lambda_3$ and where each $\lambda_i$ is a linear function of $x$, $y$, and $z$, produces a $7$-parameter family of distinct algebraic surfaces of constant breadth $2$, and the generic one of these has no nontrivial symmetries. (In fact, one only gets continuous symmetries when $p_0 = \lambda^3$ for some linear function $\lambda$.)