There are three pairs among the products that each have the same product - $ab\cdot cd$, $ac\cdot bd$, $ad\cdot bc$. After removing one of the products, two pairs remain, as well as one unpaired product. Suppose WLOG we removed $cd$.
Now, either both $ac$ and $bd$ are between $ad$ and $bc$, or both $ad$ and $bc$ are between $ac$ and $bd$. WLOG, it's the former. Then, since they're all integers, $|ad-bc|\ge 3$. If $|ad-bc|=3$, then since $n(n+3)=(n+1)(n+2)-2$ for all $n$, the products aren't equal. That leaves $|ad-bc|=4$; let $n$ be whichever of $ad$ and $bc$ is smaller. Since we have $n(n+4)=(n+1)(n+3)-3$, $ac\cdot bd < (n+1)(n+3)$ and the products $ac,bd$ must be $n+1$ and $n+2$ in some order.
Now, $n(n+4)=(n+1)(n+2)$, so $n^2+4n=n^2+3n+2$ and $n=2$. Our pairs are $(2,6)$ and $(3,4)$, leaving $5$ unpaired. The removed number $bd$ must form a pair of product $12$ with $5$, so it was $\frac{12}{5}=\boxed{2.4}$. Done.
Well, almost done. We've established what the removed number had to be, but we haven't established that it was possible at all. That just requires an example. Playing with it a bit, I find $a=2\sqrt{\frac56}$, $b=3\sqrt{\frac56}$, $c=2\sqrt{\frac65}$, and $d=\sqrt{\frac65}$ works.
Disclaimer
This question is poorly written by the author(s), since it leaves room for interpretation of the rules. The first part of this answer is based on the interpretation that "the move must be different from the last move played by them" means that apart from both player's initial moves, no player is allowed to repeat what their opponent did last move. In a second part, I will adress the reading that "the move must be different from the last move played by them" means that apart from both player's initial moves, no player is allowed to repeat what they themselves did last move. Finally, in a third part, I will adress the case where both of these apply.
Part I: no last-move copycat
Spoiler: Alice wins for $d\in \{2,3\}$, Bob wins for $d\in \{1,4,5,...\}.$
1. Implication of rules
The qoq "best" way to think about this specific game is to understand the power of position, i.e. being the person to not act first, paired with the power of moving $0$, which Bob can always do turn 1, and Alice can never repeat as per rules.
This is enough to deduce that Alice must win on turn 1, which is possible iff $d=2$ or $d=3$.
For all other values of $d$, Bob wins. Let's have a look at a few of them.
2. Discussion of first few cases
For $d=1$, Alice either overshoots with an initial $A2$ and $A3$, or prolongs her game by opening $A0$, when Bob in turn elects the play the sequence $\{A0-B0-AX\}$, forcing Alice to overshoot. (Note that from a strategical point of view, all of Alice's opening moves are optimal, as they all loose by force; practically however, $A0$ is preferable since it allows Bob to play suboptimally.)
For $d=4$, we have
$\{A3-B0-AX\},$ where Alice overshoots, or
$\{A2-B2\}$, where Bob lands, or
$\{A0-B2-A0-B2\}$ and $\{A0-B2-AX\}$. Bob wins.
For $d=5$, we have what you described, $\{A0-B0-A2-B3\}$ or $\{A0-B0-A3-B2\}.$
For $d\gt5$, the recurring scheme is that Bob can force Alice into a line described above.
F.e. for $d=6$, we have the simple $\{A3-B3\}$ for opener $A3$, $\{A2-B2-AX\}$ or $\{A2-B2-A0-B2\}$ for opener $A2$, and for opener $A0$ we have $\{A0-B3-A2-B0-AX\}$ or $\{A0-B3-A0-B3\}$. Bob wins.
For $d=7$, we either have prefix $\{A0-B0-A2-B0-\}$ and now $d=5$, with Alice to act and no $0$-move at hand, or we have prefix $\{A0-B0-A3-B0-\}$ and now $d=4$, again with Alice to act and no $A0$-move at hand. Alice may thus not open with $A0$. But opening with $A2$ brings herself directly into $d=5$, which is a win for Bob, since he can play $B0$, and opening $A3$ brings herself into $d=4$, which we saw earlier is also a win for Bob.
For $d=8$, which should be the last case of interest, since it's Alice's $d=3(+5)$, we have straight away that both $A3$ and $A2$ lead to $d=5$ and $d=6$ respectively, where Bob has a win as seen before. Again however, Alice cannot force Bob into such lines with roles reversed with $A0$, since Bob can answer with $B0$, effectively re-reversing roles.
3. Conclusion
Alice can thus never force Bob into "her" two winning cases $d=2$ and $d=3$, because Bob can prevent this by playing $B0$. In fact, Bob can continuously play $B0$ and let Alice close the distance with $A2$ and $A3$ until $4\le d\le 6$ when Bob either snaps the win or forces Alice to overshoot as seen above. This is in line with the observation that Alice must win with her initial move, otherwise Bob has a win. So in short, Bob in last position can always move when they are in a Bob's endgame line and always pass otherwise, effectively revearsing roles, or strategy-stealing.
Part II: no second-to-last-move copycat
Spoiler: Alice wins for $d\in \{2,3,4\}$ and in $(d\equiv 4\mod5).$ Bob wins for $d\in \{1,5,6,7,8,10...\}.$
This game is what the author(s) must have had in mind, since $d=94$ is indeed a win for Alice. Interestingly, it comes with a new strategic option for Bob, which is to reduce to $d\in \{2,3\}$ when Alice is prevented from finishing due to her last move having been what is required. This allows Bob to steer into wins when $(d\equiv 2 \mod5, d\gt5)$. Notably for $(d\equiv 3,4\mod 5)$ however, Alice wins.
1. Discussion of first few cases
For $d=1$, we have again $\{A0-B0-AX\}$, and Bob wins.
For $d \in \{2,3\}$, Alice wins turn $1$.
For $d=4$, Alice wins with $\{A3-B0-A0-BX\}$ where she forces Bob into overshooting.
For $d=5$, again no changes, still $\{A0-B0-A3-B2\}$ and $\{A0-B0-A2-B3\}$ for a win for Bob.
For $d=6$, we have the trivial $\{A3-B3\}$, so Alice can't open $A3$. For $A2$, we have $\{A2-B3-A0-B0-AX\}$ where Bob brings $d$ down to $1$ and forces Alice to overshoot. For $A0$ lastly, Bob repeats with $B0$, denying the revearsing of roles and forcing Alice to open into $d \in \{3,4\}$. Note that Bob didn't play $B0$ turn $2$ before, which validates the prefix $\{A0-B0-...\}$.
For $d=7$, we have $\{A3-B3-A0-B0-AX\}$ where Bob reduces to $d=1$, and $\{A2-B3-A0-B2\}$, where Bob reduces to $d=2$ at a time when Alice can't close the game with $A2$, as this was her last move. This is the above-mentioned strategical play for Bob in position, and it highlights the power of position. For the $A0$ opening, note again that Bob is safe playing $B0$, since his follow-up do not need him to play a second turn $B0$ again, validating the prefix $\{A0-B0-...\}$. So $\{A0-B0-A2-B3-A0-B2\}$ and $\{A0-B0-A3-B3-A0-B0-AX\}.$
For $d=8$, Alice wins with $A2$, reducing to $d=6$ while holding the option of $A0$ next turn. Against $A2$, if Bob plays $B2$ and reduces to $d=4$, Alice can force Bob to overshoot in sequence $\{A2-B2-A3-B0-A0-BX\}.$ Against Bob's $B3$, Alice wins immediately with $\{A2-B3-A3\}.$ Lastly, should Bob reply with $B0$, Alice is now in position to re-revearse with $\{A2-B0-A0-B2-A3-B0-A0-BX\}$ or $\{A2-B0-A0-B3-A3\}.$ So, Alice wins for $d=8$ and for all $(d\equiv 3\mod5,d \gt5)$
For $d=9$ and in fact all $(d\equiv 4\mod 5)$, Alice wins.
2. Conclusion
The interpretation that the players may not repeat their own last move, i.e. the second to last move overall, helps Bob in position, in that it opens Bob up to reduce to $d \in \{2,3\}$ exactly when Alice is forced by the rules to play a move different from what's required to land on Bob's spot. However, Alice wins for initial $(d\equiv 3,4\mod 5)$, as well as initial $d \in \{2,3\}$ where Alice wins turn $1$.
This result is in line with your assumption.
Part III: neither second-to-last-move nor last move copycat
Spoiler: Alice wins for $d\in\{2,3\}$ and for $(d\equiv 1,2 \mod 5, d\gt5)$.
It should be clear that the game is the same as the first one for $d \in \{1,2,3,4,5\}$. So, what about $d\gt5$?
For $d=6$, trivial is $\{A3-B3\}$. Against $A2$ however, Bob loses by force with all his replies, $\{A2-B3-A0-B2\},\{A2-B2-A0-B3\},\{A2-B0-A3-B2\}.$
This now highlights that starting with Bob's first move, we have repeating $3$-cycles, reducing to $d-5$ over three plies. Alice wins games where $(d \equiv 1 \mod 5, d \ge 2).$
For $d=7$, we have the same when Alice reduces to $d=4$ again, now with $A3$. We have $\{A3-B3-A0-B2\}$ and Bob overshoots, $\{A3-B2-A0-B3\}$ also overshooting, $\{A3-B0-A2-B3\}$ again overshooting. Alice wins games where $(d \equiv 2 \mod 5, d \ge 5).$
For $d=8$, Bob wins by completing to $5$ in $\{A3-B2-A0-B3\}$ and in $\{A2-B3-A0-B2-A3\}$ where Alice overshoots. Alice also loses with an initial $A0$ on the base of prefix $\{A0-B0-...\}$. Bob wins games where $(d \equiv 3 \mod 5, d \ge 5).$
For $d=9$, Bob wins with $\{A3-B3-A2-B0-AX\}$ and $\{A3-B3-A0-B2-AX\}$ against $A3$, and against $A2$ with $\{A2-B3-A0-B2-AX\}.$ The prefix case $\{A0-B0\}$ splits into the trivial tail against $A2$ with $\{A0-B0-A2-B3-A0-B2-AX\}$ and into the restricted case of $\{A0-B0-A3-B2-A0-B3-AX\}$ where Bob could not play $B3$ turn $2$ as opposed to above, but still wins. Bob wins games where $(d \equiv 4 \mod 5, d \ge 5).$
Bob also wins games where $d \equiv 0 \mod 5, d \ge 2$, since he can simply complete to $5$ or play our double pass-prefix.
2. Conclusion
Since the game turns into cycles from turn $2$, Bob can no more prevent Alice from revearsing roles; instead, Alice now has options to steer the game into favourable endgames. That is, Alice wins for $d\in\{2,3\}$ and for $(d\equiv 1,2 \mod 5, d\gt5)$, Bob wins the rest.
Best Answer
Hint: if the collections $(a_1, \dots, a_k)$ and $(b_1, \dots, b_k)$ have identical pairwise sums, then the collections $(a_1, \dots, a_k, b_1+m, \dots, b_k+m)$ and $(b_1, \dots, b_k, a_1+m, \dots, a_k+m)$ also have identical pairwise sums. (The number $m$ has to be such that $a_i \neq b_j \pm m$ for all $i,j$, so that the numbers in each collection would be different.)