I am pretty sure that it is, from what (little, so far) I understand, but...
Most of what I have read about the anomalous magnetic dipole moment and it's contribution to the electron's $g$-factor do not mention the word 'vacuum' directly....
I want to wrap my head around the concept, if possible, in addition to the math(s).....
Without knowing your background, I want to try and give you a taste without diving too much into the math, which is very complicated. I'll give for granted that you know where the Landè factor $g$ comes from and how it relates to the magnetic moment of the electron.
As per the naïve calculation, one should have $g=2$. What Schwinger showed was that, when evaluating the loop corrections to the vertex of QED, one gets a corrections to the $g$ factor.
The image shows the second order QED vertex correction. One might ask why the vertex correction should change the value of $g$ and so of the electron magnetic moment. This comes from the following (and here's some of the math, please note that the 4-momenta in the image are not the same that I'll use, the image is just an example): the most general form of the vertex function which describes the coupling between a photon and an on-shell electron is
$$\bar{u}(p^\prime)\Gamma^\mu(p^\prime, p)u(p) = \bar{u}(p^\prime)\left[f_1(q^2)\gamma^\mu+f_2(q^2)p^\mu+f_3(q^2)p^{\prime\mu}+f_4(q^2)\sigma^{\mu\nu}p_\nu+f_5(q^2)\sigma^{\mu\nu}p^\prime_\nu\right]u(p)\tag{1}\label{1}$$
where $q=p^\prime-p$ and the mass-shell relation holds $p^{\prime 2}=p^2=m^2$. Without going into much details, using conservation of the current and the Gordon identity it can be shown that equation $(\ref{1})$ is equivalent to
$$\bar{u}(p^\prime)\left[F_1(q^2)\gamma^\mu+F_2(q^2)\frac{1}{2m}i\sigma^{\mu\nu}q_\nu\right]u(p)\tag{2}\label{2}$$
where all the magnetic information is encoded in the structure function $F_2(q^2)$. Again, just to give you a taste, the calculation for the vertex correction gives a divergent integral and it can be shown that, in all orders of perturbation theory, the divergent part of the integrals is proportional to $\gamma^\mu$ and therefore we can put all the divergence in a constant $L$ that than is going to be cancelled by the corrections to the external lines
$$\Gamma^\mu(p^\prime,p) = L\gamma^\mu+\Gamma_c^\mu(p^\prime, p)$$
where $\gamma_c^\mu$ is the convergent part. In the limit $q\to 0$ the vertex function is proportional only to $\gamma^\mu$ since is the bare vertex and therefore
$$\bar{u}(p)\Gamma^\mu_c(p,p)u(p)=0$$
From $(\ref{2})$ we can see that since the term in $F_2$ is explicitly proportional to $q$, in the limit of $q\to 0$ vanishes and then $F_1(0)=L$. Expanding the RHS of $(\ref{2})$ in orders of $q^2$, we find
$$\bar{u}(p^\prime)\Gamma^\mu(p^\prime,p)u(p) = \bar{u}(p^\prime)\left[L\gamma^\mu+F_2(0)\frac{1}{2m}i\sigma^{\mu\nu}q_\nu\right]u(p)+O(q^2)$$
Because of the Ward identity, the term $L$ cancels with the correction to the external lines. Therefore, including all second order corrections, the vertex function becomes
$$ie_0\bar{u}(p^\prime)\gamma^\mu u(p)\to ie\bar{u}(p^\prime)\left[\gamma^\mu+\frac{e}{2m}F_2(0)i\sigma^{\mu\nu}q_\nu\right]u(p)+O(q^2)$$
Using again Gordon identity one finds that
$$\bar{u}(p^\prime)\Gamma^\mu(p^\prime, p)u(p) = ie\bar{u}(p^\prime)\left[\frac{(p+p^\prime)^\mu}{2m}+\frac{e}{2m}(1+F_2(0))i\sigma^{\mu\nu}q_\nu\right]+O(q^2)$$
The result obtained shows that the magnetic moment of the electron is equal to $1+F_2(0)$ Bohr's magnetons. We can therefore write the Landè $g$ factor as $$g=2(1+F_2(0))=2+O(\alpha)$$
From the full calculation it turns out that the correction to the $g$ factor is $$g=2\left(1+\frac{\alpha}{2\pi}\right)$$ where $\alpha$ is the fine structure constant.
The idee of "vacuum" as the OP says is behind all the corrections: particles emit virtual ones, effectively creating them from the vacuum (this is indeed all a consequence of perturbative calculations), which are then riabsorbed. The photon which gives the correction to the vertex is such a virtual particle. Moreover in the calculation I've said that we add the corrections to the external lines. This corrections again contain some sort of interaction between the particle in the external line and the vacuum.
That's all the reasoning behind it.
It is not really about "interaction with vacuum", it is more about "dressing particles".
The true electron is coupled permanently to the electromagnetic field oscillators, it is their part. However this coupling is treated in QED perturbatively - with a lot of difficulties due to too strong coupling being treated perturbatively.
There is a good atomic analogy, still unknown to many, see my paper here.
