About a month ago, a friend of mine taught me that there exist many sets of three positive integers $(a,b,c)$ where $a\not=b,b\not=c$ and $c\not=a$ such that each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square.
Then, I found that there exist infinitely many such sets. For example, for $$(a,b,c)=(F_{2n-1},F_{2n+1},F_{2n+3})$$ where $F_k$ is the $k$-th Fibonacci number defined by $F_0=0,F_1=1,F_{k}=F_{k-2}+F_{k-1}\ (k\ge 2)$, we have $$\color{red}{ab-1}=F_{2n-1}F_{2n+1}-1=\color{blue}{{F_{2n}}^2}$$ $$\color{red}{bc-1}=F_{2n+1}F_{2n+3}-1=\color{blue}{{F_{2n+2}}^2}$$ $$\color{red}{ca-1}=F_{2n+3}F_{2n-1}-1=F_{2n}F_{2n+2}+1=\color{blue}{{F_{2n+1}}^2}$$ $$\begin{align}\color{red}{ab-a-b+c}&=F_{2n-1}F_{2n+1}-F_{2n-1}-F_{2n+1}+F_{2n+3}\\&={F_{2n}}^2+1-F_{2n-1}+F_{2n+2}\\&={F_{2n}}^2+1+2F_{2n}\\&=\color{blue}{(F_{2n}+1)^2}\end{align}$$ $$\begin{align}\color{red}{bc-b-c+a}&=F_{2n+1}F_{2n+3}-F_{2n+1}-F_{2n+3}+F_{2n-1}\\&={F_{2n+2}}^2+1-F_{2n+3}-F_{2n}\\&={F_{2n+2}}^2+1-2F_{2n+2}\\&=\color{blue}{(F_{2n+2}-1)^2}\end{align}$$ $$\begin{align}\color{red}{ca-c-a+b}&=F_{2n+3}F_{2n-1}-F_{2n+3}-F_{2n-1}+F_{2n+1}\\&={F_{2n+1}}^2+1-F_{2n-1}-F_{2n+2}\\&={F_{2n+1}}^2+1-2F_{2n+1}\\&=\color{blue}{(F_{2n+1}-1)^2}\end{align}$$
I have also found many other such sets which are not included in this Fibonacci family, and...
Interestingly enough, for every set $(a,b,c)$ I found, each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is also a perfect square.
For example, for $(a,b,c)=(F_{2n-1},F_{2n+1},F_{2n+3})$, we have $$\begin{align}\color{red}{ab+a+b-c}&=F_{2n-1}F_{2n+1}+F_{2n-1}+F_{2n+1}-F_{2n+3}\\&={F_{2n}}^2+1+F_{2n-1}-F_{2n+2}\\&={F_{2n}}^2+1-2F_{2n}\\&=\color{blue}{(F_{2n}-1)^2}\end{align}$$ $$\begin{align}\color{red}{bc+b+c-a}&=F_{2n+1}F_{2n+3}+F_{2n+1}+F_{2n+3}-F_{2n-1}\\&={F_{2n+2}}^2+1+F_{2n+3}+F_{2n}\\&={F_{2n+2}}^2+1+2F_{2n+2}\\&=\color{blue}{(F_{2n+2}+1)^2}\end{align}$$ $$\begin{align}\color{red}{ca+c+a-b}&=F_{2n+3}F_{2n-1}+F_{2n+3}+F_{2n-1}-F_{2n+1}\\&={F_{2n+1}}^2+1+F_{2n-1}+F_{2n+2}\\&={F_{2n+1}}^2+1+2F_{2n+1}\\&=\color{blue}{(F_{2n+1}+1)^2}\end{align}$$
So, here is my question.
Question : Are the following propositions true?
Proposition 1 (False) : For positive integers $(a,b,c)$ such that $a\not=b,b\not=c$ and $c\not=a$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Proposition 2 (True) : For positive integers $(a,b,c)$ such that $\color{red}{1\le} a\lt b\lt c$ and $ab+a+b-c\ge 0$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Proposition 3 : For positive integers $(a,b,c)$ such that $\color{red}{2\le} a\lt b\lt c$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Proposition 4 (True) : For positive integers $(a,b,c)$ such that $\color{red}{1\le} a\lt b\lt c$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab\mp (a+ b-c),\ bc\mp (b+ c- a),\ ca\mp (c+ a-b)$$ is a perfect square, then $c=a+b+2\sqrt{ab-1}$ holds.
Added : Proposition 1 is false because of $(a,b,c)=(1,5,65)$ found by Oleg567. Then, I added proposition 2 and 3. Also, I added proposition 4 (see Tito Piezas III's answer).
Can anyone help?
Update : I crossposted to MO.
Added : Proposition 2 and 4 are true (see my answer for the details). However, we still don't know if proposition 3 is true.
(Update.) A quick search of,
$$\begin{aligned} ab-1\;&=d^2\\ ac-1\;&=e^2\\ bc-1\;&=f^2 \end{aligned}$$
with $1<a<b<c$ gives,
$$2,\;5,\;13\\2,\;13,\;25\\2,\;25,\;41\\5,\;10,\;29\\5,\;13,\;34\\5,\;29,\;58$$ Certain patterns are immediately apparent. One of which is that the first 100 solutions with $c$ below a bound (and even higher) ALL satisfy,
$$(a+b+c)^2 = 2(a^2+b^2+c^2+2)\tag1$$
Equivalently (by solving for $c$),
$$c=a+b\pm2\sqrt{ab-1}=a+b\pm 2d$$
This has implications for the original proposition by the OP. (The first solution that does not satisfy the relation is $2,\,5,\,925.$)
I. Assumption 1.
There is a missing assumption in the Question section. It suffices to find two integers $a,b$ such that,
$$ab-1 = d^2\tag{2a}$$
then define $c$ as,
$$\color{blue}{c = a + b - 2 d}\tag{2b}$$
and these two conditions are sufficient to ensure that the following nine expressions necessarily are squares,
$$\begin{aligned} a b-1\; &= d^2\\ a c-1\; &= (a - d)^2\\ b c-1 \; &= (b - d)^2\\ a b - a - b + c \; &= (1 - d)^2\\ b c + a - b - c \; &= (1 - b + d)^2\\ a c - a + b - c \; &= (1 - a + d)^2\\ a b + a + b - c \; &= (1 + d)^2\\ b c - a + b + c \; &= (1 + b - d)^2\\ a c + a - b + c \; &= (1 + a - d)^2 \end{aligned}$$
There are broad families to the 2 conditions. One is $a,\,b,\,c = x-ny,\;x+ny,\;2(x+my)$ where $x,y$ solve the Pell equation,
$$x^2-(m^2+n^2)y^2 = 1$$
This covers the Fibonacci family by the OP which was just the case $m,n= 2,1.$ (See also this similar post.)
II. Assumption 2.
For the counter-examples found by Oleg567 to the original proposition, if,
$$a=1,\quad \text{so}\quad b,c = p^2+1,\;q^2+1\tag3$$
$$\begin{aligned} a b-1\; &= p^2\\ a c-1\; &= q^2\\ \color{brown}{b c-1} \; &= p^2+q^2+p^2q^2\\ a b - a - b + c \; &= q^2\\ b c + a - b - c \; &= p^2q^2\\ a c - a + b - c \; &= p^2\\ \end{aligned}$$
If $q=2p^2$, then $\color{brown}{b c-1}$ also becomes a square. However, the remaining three numbers will not necessarily be so.
III. Summary
It is asked, "For $2\leq a < b < c$, if the six numbers are perfect squares, then are the three (remaining) numbers perfect squares?".
I assume that the six expressions being squares are not sufficient, by themselves, to ensure that the remaining three expressions will be also squares (like Oleg's counter-examples). The fact that for $a>1$ the remaining three numbers were squares seem to be more due to the additional condition $(1)$.
But it would be interesting to find $a,b,c$ with $a>1$ such that only the first six numbers are squares.