$Av_0 \wedge Av_{-1} \wedge Av_{-2} \wedge \cdots$ does make sense, but it cannot be "multiplied out" in finitely many steps, so the usual definition does not work. Instead, one needs to define semiinfinite wedges slightly more generally. Here is one such definition (from §3.14.3 in my notes from Etingof's class on infinite-dimensional Lie algebras; but this section is my work, so don't be surprised if there is something much simpler):

**Definition 1.** **(a)** A family $\left( x_{i}\right)
_{i\in\mathbb{Z}}$ of elements of some additive group is
said to be *semiinfinite* if every sufficiently high $i\in\mathbb{Z}$
satisfies $x_{i}=0$.

**(b)** Let $\widehat{V}$ be the vector subspace $\left\{ v\in
\mathbb{C}^{\mathbb{Z}}\ \mid\ v\text{ is semiinfinite}\right\} $ of
$\mathbb{C}^{\mathbb{Z}}$. Every element $\left( x_{i}\right) _{i\in\mathbb{Z}}$ of
$\widehat{V}$ is identified with the column vector $\left(\ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, \ldots\right)^T$.

**(c)** Let $V$ be the vector subspace $\left\{ v\in
\mathbb{C}^{\mathbb{Z}}\ \mid\ \text{all but finitely many entries of } v\text{ are zero}\right\} $ of $\mathbb{C}^{\mathbb{Z}}$. Clearly, $V \subset \widehat{V} \subseteq \mathbb{C}^{\mathbb{Z}}$.

**(d)** Let $\left(v_i\right)_{i \in \mathbb{Z}}$ be the standard basis of the $\mathbb{C}$-vector space $V$; that is, $v_i$ is the vector whose $i$-th coordinate is $1$ while all its other coordinates are $0$.

Notice that my $V$ is probably not your $V$.

**Definition 2.** Let $\ell\in\mathbb{Z}$. Let $\pi_{\ell}:\widehat{V}\rightarrow V$ be the linear map which sends every $\left(
x_{i}\right) _{i\in\mathbb{Z}}\in\widehat{V}$ to $\left( \left\{
\begin{array}
[c]{c}%
x_{i}\text{, if }i\geq\ell;\\
0\text{, if }i<\ell
\end{array}
\right. \right) _{i\in\mathbb{Z}}\in V$. (It is very easy to see that this
map $\pi_{\ell}$ is well-defined.)

I denote the $m$-th semiinfinite wedge space (which you probably call $F^{(m)} V$) by $\wedge^{\dfrac{\infty}{2},m}V$.

I assume that you know how to define $b_0 \wedge b_1 \wedge b_2 \wedge \cdots \in \wedge^{\dfrac{\infty}{2},m} V$ for every $m \in \mathbb{N}$ and every sequence $b_{0},b_{1},b_{2},\ldots$ of vectors in $V$ with the property that
$$
b_{i} =v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all
sufficiently large }i.
$$
But now we want a more general definition:

**Definition 3.** Let $m\in\mathbb{Z}$. Let $b_{0},b_{1},b_{2},\ldots$ be
vectors in $\widehat{V}$ which satisfy
\begin{equation}
\pi_{m-i}\left( b_{i}\right) =v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all
sufficiently large }i.
\label{darij1.def3.leading-terms}
\tag{1}
\end{equation}
(Roughly speaking, this means that for each sufficiently large $i$, the "leading term" of $b_i$ is $v_{m-i}$.)
Define an element $b_{0}\wedge b_{1}\wedge b_{2}\wedge\cdots$ of $\wedge
^{\dfrac{\infty}{2},m}V$ as follows: Pick some $N\in\mathbb{N}$ such that
every $i>N$ satisfies $\pi_{m-i}\left( b_{i}\right) =v_{m-i}$. (Such an $N$
exists, because of \eqref{darij1.def3.leading-terms}.) Then, we define $b_{0}\wedge b_{1}\wedge b_{2}\wedge\ldots$ to be the element
$$
\begin{align}
& \pi_{m-N}\left( b_{0}\right) \wedge\pi_{m-N}\left( b_{1}\right)
\wedge\ldots\wedge\pi_{m-N}\left( b_{N}\right) \wedge v_{m-N-1}\wedge
v_{m-N-2}\wedge v_{m-N-3}\wedge\ldots \\
& \in\wedge^{\dfrac{\infty}{2},m}V.
\end{align}
$$
This element does not depend on the choice of $N$ (this is easy to see). Hence, $b_{0}\wedge b_{1}\wedge b_{2}\wedge\cdots$ is well-defined.

According to Definition 3, the element $Av_0 \wedge Av_{-1} \wedge Av_{-2} \wedge \cdots$ is well-defined for every $A$ for which $\pi_{-i}\left( Av_{-i}\right) =v_{-i}$ for all sufficiently large $i$ (here, I have set $m=0$, because this is the case you care about). This is clearly the case for $A \in \overline{\operatorname{GL}}_\infty$. (Actually, the invertibility of $A$ is not needed.)

In your example, **every** $i$ satisfies $\pi_{-i}\left( Av_{i}\right) =v_{-i}$, and thus you have $Av_0 \wedge Av_{-1} \wedge Av_{-2} \wedge \cdots = v_{-0} \wedge v_{-1} \wedge v_{-2} \wedge \cdots$. This generally holds when $A$ is an upper-triangular matrix with $1$'s on its main diagonal.