The answer is 'no', though I don't know an easy way to see this without doing an explicit calculation. Here is where to look though, if you want to do the calculation yourself:

Things work out a bit better if one uses indeterminates $z = x+y$ and $w = x-y$. Then one has an expansion
$$
\det\bigl(x{\cdot}I + y{\cdot}\star + A\bigr) = \sum_{0\le i,j\le 3} P_{ij}(A)\,z^{3-i}w^{3-j}
$$
where $P_{ij}(A)$ is a polynomial in $A$ of degree $i{+}j$.

One finds that $P_{00}=1$. Meanwhile, the polynomials $P_{10}$, $P_{01}$, $P_{20}$, $P_{11}$, $P_{02}$, $P_{30}$, $P_{21}$, $P_{12}$, and $P_{03}$ are independent and generate all of the polynomial invariants of degree $3$ or less. Moreover, the three quartic polynomials $P_{3,1}$, $P_{2,2}$, and $P_{1,3}$ are algebraically independent over the ring generated by the lower degree $P_{ij}$.

However, the $P_{ij}$ with $i+j\le 4$ do not generate all of the invariant polynomials of degree $4$. There is one further quartic polynomial invariant, say $Q$, that is not a polynomial in the $P_{ij}$ with $i+j\le 4$. Since the remaining three coefficients $P_{3,2}$, $P_{2,3}$ and $P_{3,3}$ are all of degree 5 or higher, $Q$ does not lie in the ring generated by the $P_{ij}$.

**Added detail:** $\mathrm{SO}(4)$ preserves the splitting $\Lambda^2(\mathbb{R}^4)= \Lambda^2_+(\mathbb{R}^4)\oplus\Lambda^2_-(\mathbb{R}^4)$ into self-dual and anti-selfdual forms, i.e., the split into the eigenspaces of $\star$, which has trace-zero and square $\star^2 = I$. It follows that we can write block decompositions
$$
\star = \begin{pmatrix} I_3&0_3\\0_3&-I_3\end{pmatrix}
\quad\text{and}\quad
A = \begin{pmatrix} a&b\\b^T&c\end{pmatrix},
$$
where $a=a^T$, $b$ and $c=c^T$ are $3$-by-$3$ matrices. The action of $\mathrm{SO}(4)$ on $A$ can be described as
$$
g\cdot A = \begin{pmatrix} g_+ag_+^T&g_+bg_-^T\\g_-b^Tg_+^T&g_-cg_-^T\end{pmatrix}
$$
where $g_\pm:\mathrm{SO}(4)\to\mathrm{SO}(3)$ are homomorphisms such that $(g_+,g_-):\mathrm{SO}(4)\to\mathrm{SO}(3)\times\mathrm{SO}(3)$ is surjective, with kernel $\{\pm I_4\}$. We can write
$$
\det(x{\cdot}I_6 + y{\cdot}\star + A)
= \det\begin{pmatrix} zI_3+a & b\\ b^T& wI_3 + c\end{pmatrix}
= \sum_{0\le i,j\le 3} P_{ij}(A)\,z^{3-i}w^{3-j},
$$
and we can refine our notation by writing $P_{ij}(A) = P_{ij}(a,b,c)$.

Now, because of the way $a$, $b$, and $c$ transform under the action of $\mathrm{SO}(4)$, we see, for example, that $\mathrm{tr}(a^k)$ and $\mathrm{tr}(c^l)$ are invariant polynomials, as is $\mathrm{tr}(bb^T)$ (though $\mathrm{tr}(b)$ is not), and, indeed, we see, by inspection, that, for example,
$$
P_{10} = \mathrm{tr}(a)\quad\text{and}\quad P_{01} = \mathrm{tr}(c),
$$
and a little reflection shows that
$$
P_{20} = \tfrac12\bigl(\mathrm{tr}(a)^2-\mathrm{tr}(a^2)\bigr)
\quad\text{and}\quad
P_{02} = \tfrac12\bigl(\mathrm{tr}(c)^2-\mathrm{tr}(c^2)\bigr)
$$
while it's not hard to see that
$$
P_{11} = \mathrm{tr}(a)\mathrm{tr}(c) - \mathrm{tr}(bb^T).
$$
To go further, imagine that we have an alphabet of 4 letters, $a$, $b$, $b^T$, and $c$, and we declare a valid `word' to be string of these letters subject to the following rules: $a$ can be followed only by $a$ or $b$, $b$ can only be followed by $c$ or $b^T$, $c$ can only be followed by $c$ or $b^T$, $b^T$ can only be followed by $a$ or $b$, and finally, that the number of $b$s in the word is the same as the number of $b^T$s. Thus, for example, $a^4$, $abb^T$, and $abcb^T$ are valid words, but $abc$ is not valid. What one sees that is a vaild word $w$, when interpreted as a product of matrices, transforms under $\mathrm{SO}(4)$ into $g_\pm w g_\pm^T$, and hence $\mathrm{tr}(w)$ is an invariant polynomial for any valid word.

Now we can go on: For example
$P_{30} = \det(a)$ which is a weighted homogeneous polynomial in $\mathrm{tr}(a)$, $\mathrm{tr}(a^2)$, and $\mathrm{tr}(a^3)$, and similarly with $P_{03}= \det(c)$. Moreover, one finds that
$$
P_{21} = P_{20}P_{01}-P_{10}(P_{10}P_{01}-P_{11})
+\mathrm{tr}(abb^T)
$$
with a similar formula for $P_{12}$. Thus, all of the polynomials $P_{ij}$ for $0<i+j\le 3$ are expressed as polynomials in the 9 (independent) invariants
$$
\mathrm{tr}(a),\mathrm{tr}(c),\ \mathrm{tr}(a^2),\mathrm{tr}(c^2),\mathrm{tr}(bb^T),\ \mathrm{tr}(a^3),\mathrm{tr}(abb^T),\mathrm{tr}(cb^Tb), \mathrm{tr}(c^3),
$$
and conversely. These are all of the invariants of degree $3$ or less. (Of course, $\mathrm{tr}(a^4)$ is a polynomial in these. Not all valid words yield an invariant that cannot be expressed as a polynomial in invariants of lower degree.)

But now, one sees where the discrepancy comes in. The quartic polynomials $P_{31}$, $P_{22}$, $P_{13}$ can be expressed in terms of the 'word-trace' invariants, but there are ony three of them, whereas there are *four* new quartic word-trace invariants:
$$
\mathrm{tr}(aabb^T),\ \mathrm{tr}(bb^Tbb^T),\ \mathrm{tr}(ccb^Tb),\ \mathrm{tr}(abcb^T)
$$
and it is not hard to see that no nontrivial linear combination of them can be written as a polynomial in the word-trace invariants of lower degree.

In fact, the situation gets worse: There are only two $P_{ij}$ of degree 5, but there are *four* new word-trace invariants of degree 5, and $P_{33}$ is the only one of degree 6 while there are *two* new word-trace invariants of degree 6.