jmc

We can directly compute $$\left(\frac{3}{4}+\frac{3}{4}i\right)z=\left(\frac{3}{4}+\frac{3}{4}i\right)(x+iy)=\frac{3(x-y)}{4}+\frac{3(x+y)}{4}\cdot i.$$This number is in $S$ if and only if $-1\le \frac{3(x-y)}{4}\le 1$ and at the same time $-1\le \frac{3(x+y)}{4}\le 1$. This simplifies to $|x-y|\le \frac{4}{3}$ and $|x+y|\le \frac{4}{3}$.

Let $T=\{x+iy:|x-y|\le \frac{4}{3}\text{and}|x+y|\le \frac{4}{3}\}$, and let $[X]$ denote the area of the region $X$. Then, the probability we seek is $\frac{[S\cap T]}{[S]}=\frac{[S\cap T]}{4}$. All we need to do is to compute the area of the intersection of $S$ and $T$. It is easiest to do this graphically:



Coordinate axes are dashed, $S$ is shown in red, $T$ in green and their intersection is yellow. The intersections of the boundary of $S$ and $T$ are obviously at $(\pm 1,\pm 1/3)$ and at $(\pm 1/3,\pm 1)$.

Hence, each of the four red triangles is an isosceles right triangle with legs of length $\frac{2}{3}$, and the area of a single red triangle is $\frac{1}{2}\cdot {\left(\frac{2}{3}\right)}^2=\frac{2}{9}$. Then, the area of all four is $\frac{8}{9}$, and therefore the area of $S\cap T$ is $4-\frac{8}{9}$. Thus, the probability we seek is $\frac{[S\cap T]}{4}=\frac{4-\frac{8}{9}}{4}=1-\frac{2}{9}=\boxed{\frac{7}{9}}$.