Alice has a winning strategy. By the following lemma, as long as Alice makes $A_7 \in \{1,3,7,9\}$, Alice will necessarily win; and since Bob can block at most three of these, Alice must win.

Lemma
If $0 < a < 10^7$ and $\text{gcd}(a, 10) = 1$, then $x^7 \equiv a \pmod{10^7}$ has a solution.

Proof
We first prove:

Claim. If $\text{gcd}(m, 10) = \text{gcd}(n, 10) = 1$ and $10^7|m^7 - n^7$, then $10^7|m - n$.

Pf of Claim. Note that
$$m^7 - n^7 = (m-n)(m^6 + m^5n + m^4n^2 + m^3n^3 + m^2n^4 + mn^5 + n^6).$$
Therefore $\text{gcd}(m-n, \frac{m^7-n^7}{m-n}) = \text{gcd}(m-n, 7n^6)$. Also note that $7n^6$ is not a multiple of $2$ or $5$, so $2^7|m-n$ or $2^7|(m^7-n^7)/(m-n)$, and $5^7|m-n$ or $5^7|(m^7-n^7)/(m-n)$. We now prove that both "or" statements must hold in the former way.

1. Since $m$ and $n$ must both be odd, and the two factors cannot have a common factor, it follows that $2^7|m-n$.
2. Note that if $m \equiv 1,2,3,4 \pmod 5$, then $m^7 \equiv 1,3,2,4 \pmod 5$. Substituting this in, it is easy to see that $5^7|m^7-n^7$; and $2^7|m-n \Rightarrow 5^7|m-n$.

Combining the above, the Claim is proved.

Returning to the original problem, by the Claim, the reduced residue system modulo $10^7$, after being raised to the seventh power, remains a reduced residue system, hence $x^7 \equiv a(\text{mod } 10^7)$ always has a solution.