Given $$P(I\mid J)=0.7$$$$P(I^c\mid J^c)=0.3$$ does the following bit of working out show that $I$ and $J$ are not independent events?
$$P(I\mid J)P(J)=P(I \cap J)=0.7P(J)$$
$$P(I^c\mid J^c)P(J^c)=P(I^c \cap J^c)=0.3P(J^c)$$
Since $P(I^c \cap J^c)=1-P(I \cap J)$,
and $P(J^c)=1-P(J)$
$$1-0.7P(J)=0.3(1-P(J))$$
$$1-0.7P(J)=0.3-0.3P(J)$$
$$0.4P(J)=0.7 \rightarrow P(J)=7/4,$$ which is silly.
I feel like there's a better way to show that the two are not independent though...
Short solution: Assume $I,J$ independent in the space $\Omega$. Then the two $\sigma$-algebras generated by $I$, which is $\{\emptyset, I, I^c,\Omega\}$, and $J$, which is $\{\emptyset, J, J^c,\Omega\}$, are also independent, so $P(I|J)=P(I)$, and $P(I^c|J^c)=P(I^c)$, but the sum $P(I)+P(I^c)\ne 1$, contradiction.
Longer solution that follows the thoughts in the OP.
Let us use variables for $P(I), P(J), P(I\cap J)$ to have a simple road through algebraic manipulations. Let us assume $I,J$ independent, and try to get a contradiction. Notations and given relations, under this assumption: $$ \begin{aligned} P(I) &=: a\in[0,1]\ ,\\ P(J) &=: b\in[0,1]\ ,\\ P(I\cap J) &=ab=: c\in[0,a\wedge b]\ ,\\ P(I\cup J) &=a+b-c\ ,\\ P(I^c\cap J^c)&=P(\ (I\color{red}{\cup} J)^c\ )\\ &=1-P(I\cup J) \\ &=1-a-b+c \\ &=1-a-b+ab \\ &=(1-a)(1-b)\ ,\\[2mm] 0.7&=P(I|J)=\frac cb=a\ ,\\ 0.4&=P(I^c|J^c)=\frac {1-a-b+c}{1-b}=1-a\ , \end{aligned} $$ contradiction.