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The answer is: of course not. Indeed, take any real $a\ne0$, any real $b$, and any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the real and imaginary parts of the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=Q(as)Q(bt)\,\ln\big(Q(as)Q(bt)\big)\in\mathbb C$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which it does not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $a\,\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.

The answer is: of course not. Indeed, take any real $a\ne0$, any real $b$, and any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the real and imaginary parts of the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=Q(as)Q(bt)\,\ln\big(Q(as)Q(bt)\big)\in\mathbb C$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which the do not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $a\,\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.

The answer is: of course not. Indeed, let $a=1$, $b=0$, and take any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=\frac{Q(s)}2\,\ln\frac{Q(s)}2$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which it does not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.

The answer is: of course not. Indeed, take any real $a\ne0$, any real $b$, and any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the real and imaginary parts of the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=Q(as)Q(bt)\,\ln\big(Q(as)Q(bt)\big)\in\mathbb C$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which it does not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $a\,\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.

The answer is: of course not. Indeed, let $a=1$, $b=0$, and take any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=\frac{Q(s)}2\,\ln\frac{Q(s)}2$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which it does not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $s$ is $0$ everywhere.

The answer is: of course not. Indeed, let $a=1$, $b=0$, and take any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=\frac{Q(s)}2\,\ln\frac{Q(s)}2$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which it does not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.