The set $A$ may or may not be closed, depending on the function $f$.

For instance, if $f$ is of the form $f(x,y) = g(x) h(y)$ for continuous $g,h \in C([0,1])$, then $A = A_f$ is closed: if we let $I_f(\mu) = \int f\,d(\mu \times \mu)$ then by Fubini's theorem we have $$I_f(\mu) = \left(\int g\,d\mu \right) \left(\int h\,d\mu\right)$$
which is the product of two weak-* continuous functions. Hence $I_f$ is weak-* continuous and so $A_f = \{I_f \le 1\}$ is closed. A similar argument applies if $f$ is a finite sum of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$. Note that by the Stone-Weierstrass theorem, the set of functions $f$ with this form is dense in $C([0,1]^2)$.

However, there also exist $f$ for which $A_f$ is not closed; indeed, $A_f$ can be dense. I will follow a similar construction to that used in https://mathoverflow.net/a/211173/4832, based on a suggestion from Anthony Quas. For convenience, I will work on $[-1,1]$ instead of $[0,1]$.

Let $f \in C([-1,1]^2)$ be the function
$$f(x,y) = \begin{cases} 1, & x < 0 \\ e^{xy}, & x \ge 0 \end{cases}$$
which is clearly continuous.

For later use, let $F \subset C([-1,1])$ be the set of all those functions which are constant on $[-1,0]$. Note that $F$ is a closed linear subspace of $C([-1,1])$ with infinite codimension, and that $f(\cdot, y) \in F$ for each $y \in [-1,1]$.

A key fact is that the functions $\{f(\cdot, y) : y \in [-1,1]\}$ are linearly independent in $C([-1,1])$. This follows from an observation by Daniel Fischer, which I'll repeat here. If there should exist distinct $y_1, \dots, y_m$ and constants $b_1, \dots, b_m$, not all zero, such that $\sum b_j f(x,y_j) = 0$ for all $x$, then in particular, taking $x_i = (i-1)/m \in [0,1]$ for $i = 1,\dots, m$, we would have that the $m \times m$ matrix $T$ with $T_{ij} = f(x_i, y_j)$ is singular, since $b = (b_1, \dots, b_m)^t$ is in its null space. But $T_{ij} = e^{x_i y_j} = (e^{y_j/m})^{i-1}$ so that $T^t$ is a Vandermonde matrix, hence nonsingular.

Now I will show that in fact $A_f$ is dense.

Let $\mu_0 \in M([-1,1])$ be any signed measure. An arbitrary basic open neighborhood $U$ of $\mu_0$ is of the form
$$U = \bigcap_{i=1}^n \left\{\mu : \left| \int g_i\,d\mu - \int g_i\,d\mu_0 \right| < \epsilon\right\}$$
for some $\epsilon > 0$ and some $g_1, \dots, g_n \in C([-1,1])$. For brevity, set $a_i = \int g_i\,d\mu_0$. I will construct a measure $\mu$ with $\int g_i\,d\mu = a_i$ for each $i$ and $I_f(\mu) \le 1$, so that $\mu \in A_f \cap U$.

To do so, we can suppose without loss of generality that the functions $g_1, \dots, g_n$ are linearly independent in $C([-1,1])$. (If we show that $\int g_{i_j}\,d\mu = a_{i_j}$ for some $g_{i_1}, \dots, g_{i_m}$, then the same must hold for any other $g_i$ which is in their linear span, since the map $g \mapsto \int g\,d\mu_0$ is linear.) Let $E \subset C([-1,1])$ be the linear span of $g_1, \dots, g_n$.

Since, as argued above, the functions $\{f(\cdot, y) : y \in [-1,1]\}$ are linearly independent, given any finite dimensional subspace $E'$ of $C([-1,1])$ we can choose $y \in [-1,1]$, or even $y \in [-1,0]$, so that $f(\cdot, y) \notin E'$. As such, we can choose distinct $y_1, \dots, y_{n+2} \in [-1,0]$ such that
$$\{g_1, \dots, g_n, f(\cdot, y_1), \dots, f(\cdot, y_{n+2})\}$$ are linearly independent.

Now consider the linear map $L : C([-1,1]) \to \mathbb{R}^{n+2}$ defined by $L(g) = (g(y_1), \dots, g(y_{n+2}))$. Note that $\dim L(E) \le n$ and $\dim L(F) = 1$ (recall that $y_1, \dots, y_{n+2} \in [-1,0]$, so that $L(F)$ is just the constant vectors). So $\dim L(E+F) \le n+1$ and thus we may choose $(c_1, \dots, c_{n+2}) \in \mathbb{R}^{n+2} \setminus L(E+F)$.

By the linear independence of the functions $g_i$ and $f(x, y_j)$, using the Hahn-Banach and Riesz representation theorems, we can find a signed measure $\mu_1 \in M([-1,1])$ such that
$$\begin{align*}\int g_i(x)\, \mu_1(dx) &= a_i, && i = 1,\dots, n \\ \int f(x, y_j)\, \mu_1(dx) &= c_j, && j = 1, \dots, n+2.\end{align*}.$$
By construction $\mu_1 \in U$. If it should happen that $I_f(\mu_1) \le 1$, then $\mu_1 \in A_f$ and we are done.

Otherwise, let $h(y) = \int f(x,y) \,\mu_1(dx)$ which is a continuous function thanks to the dominated convergence theorem. Note that by construction, $h(y_j) = c_j$; in particular, $h \notin E+F$. Now $E$ is finite dimensional and $F$ is closed with infinite codimension, so $E+F$ is a proper closed subspace of $C([-1,1])$. Thus by Hahn-Banach and Riesz again, we can find a signed measure $\mu_2$ with $\int g\,d\mu_2 = 0$ for all $g \in E+F$, and $\int h\,d\mu = -I_f(\mu_1)$. I claim that $\mu = \mu_1 + \mu_2$ is the desired measure.

First, for the functions $g_i$ that define $U$, we have $g_i \in E$ and so $\int g_i\,d\mu_2 = 0$. Hence $\int g_i\,d\mu = \int g_i\,d\mu_1 = a_i$ so $\mu \in U$.

It remains to show that $I_f(\mu) \le 1$; in fact I claim that $I_f(\mu) = 0$. Expanding $I_f(\mu)$ in terms of $\mu_1, \mu_2$ and using Fubini's theorem, we have
$$I_f(\mu) = I_{11} + I_{12} + I_{21} + I_{22}$$
where
$$I_{ij} = \int f(x,y) \,\mu_i(dx) \mu_j(dy).$$
Now since $f(\cdot, y) \in F$ for each $y$, we have $\int f(x,y) \,\mu_2(dx) = 0$ for every $y$. Hence $I_{21} = I_{22} = 0$. We have $I_{11} = I_f(\mu_1)$ by definition, and
$$I_{12} = \int f(x,y)\, \mu_1(dx) \mu_2(dy) = \int h(y)\,\mu_2(dy) = -I_f(\mu_1)$$
by construction. Hence $I_f(\mu) = 0$ as desired.