You have the right general idea. However, rather than your $a=(b-1)^2$, which I also wasn't able to finish with, the following uses $a=b^2-1$ instead. First, we have
$$a! + (a+2)! = a! + a!(a+1)(a+2) = a!(1 + (a+1)(a+2)) = a!(a^2 + 3a + 3) \tag{1}\label{eq1A}$$
$$(a+2\lfloor \sqrt{a}\rfloor)! = a!(a+1)(a+2)\ldots(a+2\lfloor \sqrt{a}\rfloor) = a!\prod_{k=1}^{2\lfloor \sqrt{a}\rfloor}(a+k) \tag{2}\label{eq2A}$$
Thus, since $a! \neq 0$, the divisibility condition becomes
$$a!(a^2 + 3a + 3) \mid a!\prod_{j=1}^{2\lfloor \sqrt{a}\rfloor}(a+j) \;\;\to\;\; (a^2 + 3a + 3)\mid \prod_{j=1}^{2\lfloor \sqrt{a}\rfloor}(a+j) \tag{3}\label{eq3A}$$
as marty cohen's comment indicates.
Next, consider we have for some positive integer $j$ that
$$a + j \mid a^2 + 3a + 3 \tag{4}\label{eq4A}$$
Letting $a + j = k \;\to\; a \equiv -j \pmod{k}$, then from \eqref{eq4A} we have, for some positive integer $m$, that
$$a^2 + 3a + 3 \equiv (-j)^2 + 3(-j) + 3 \equiv 0 \pmod{k} \;\;\to\;\; j^2 - 3j + 3 = mk \tag{5}\label{eq5A}$$
With $m = 1$, we get
$$j^2 - 3j + 3 = a + j \;\;\to\;\; a = j^2 - 4j + 3 = (j - 2)^2 - 1 \tag{6}\label{eq6A}$$
Letting $b = j - 2$, then
$$a = b^2 - 1 \;\;\to\;\; 2\lfloor \sqrt{a}\rfloor = 2(b-1) = 2b - 2 \tag{7}\label{eq7A}$$
and
$$\begin{equation}\begin{aligned}
a^2 + 3a + 3 & = (b^2-1)^2 + 3(b^2 - 1) + 3 \\
& = b^4 - 2b^2 + 1 + 3b^2 - 3 + 3 \\
& = b^4 + b^2 + 1 \\
& = (b^2 + b + 1)(b^2 - b + 1) \\
& = ((b^2 - 1) + b + 2)((b^2 - 1) - b + 2) \\
& = (a + b + 2)(a - b + 2)
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
For $b \ge 4$, we have $b + 2 \le 2b - 2$, so $a + b + 2$ is a factor of $\prod_{j=1}^{2\lfloor \sqrt{a}\rfloor}(a+j)$ (with $j=b+2$ in \eqref{eq4A} also indicating this). As for the $b^2 - b + 1$ factor, note this answer of $x^2-x+1$ has a root $\!\bmod p\,$ for infinitely many primes $p$ proves that, for every prime $p \equiv 1 \pmod{3}$, there's a positive integer value of $b$ where $p \mid b^2 - b + 1$. As that answer shows, this comes from $(2b-1)^2\equiv -3\pmod{p}$. Let $b_0$ be a solution of this with $1 \lt b_0 \lt p-1$. Since $2b_0-1\not\equiv 0,\pm 1\pmod{p}$, then if $2b_0 - 1 \gt p + 1$, choose $b=b_0$ since we have $2b - 2 \gt p$. Otherwise, $2b_0 - 1 \lt p - 1 \;\to\; -(2b_0 - 1) \gt -p + 1 \;\to\; 2p - (2b_0 - 1) \gt p + 1$, so $2(p-b_0+1)- 2 \gt p$. Since $(2p - (2b_0 - 1))^2\equiv(2b_0 - 1)^2\equiv-3\pmod{p}$ is also a solution, we can choose $b = p-b_0+1$ (where $1 \lt b \lt p - 1$ is also true) in this case.
Note that, in both cases, we have $p \mid b^2 - b + 1$, so $\frac{b^2 - b + 1}{p}$ is an integer. Also, with $b^2 - b + 1 \lt b^2$ and $b \lt p \;\to\; b^2 \lt p^2$, we have $\frac{b^2 - b + 1}{p} \lt p \lt 2b - 2$. Thus, since for all positive integers $n$, the product of any at least $n$ consecutive integers always has at least one factor of $n$, there exists $1 \le j_1,j_2 \le 2\lfloor\sqrt{a}\rfloor - 1$ (note we can reduce this upper value by $1$) where $p \mid a + j_1$ and $\frac{b^2 - b + 1}{p} \mid a + j_2$.
The only thing left to consider is if there is any overlap of a factor $\gt 1$ between $b^2 + b + 1$ and $b^2 - b + 1$. However, any such factor must divide $(b^2 + b + 1) - (b^2 - b + 1) = 2b$, but $b^2 \pm b + 1$ are odd and $\gcd(b, b^2 \pm b + 1) = 1$.
This shows that, for each of the infinitely many primes where $p \equiv 1 \pmod{3}$ (e.g., from Dirichlet's theorem, as indicated in the previously linked answer), there's a corresponding $a$ where \eqref{eq3A} is true (e.g., for $p=7$, we have $b=5$ and $a=24$ works).
