Let $A$ be a nonzero real number and let $B$ be a nonreal complex number. Let $z$ be a complex number. Let $f(z)$ and $g(z)$ be non-constant functions defined for all complex numbers $z$ and satisfy
$f(z+A)=f(z),$ $g(z+B)=g(z)$
and $B$ is not a period of $f$ and $A$ is not a period of $g$.
Is it possible that $g(z+f(z))$ is periodic ?
Yes. There are functions $\theta(z)$, with a 1-cyclic real period, and $g(z), \;h(z)$, are complex periodic functions where $h(z)=g(z+\theta(z))$ and $g(z)$ has a different complex period than $h(z)$.
I generate an example solution below, using complex dynamics. Consider iterating the function $f(z)=z^2+z-0.01$ Also, we could use other small negative values other than -0.01, to generate two fixed points.
$f(z)$ has two fixed points, -0.1 and 0.1, where $f(-0.1)=-0.1\;\;\;\;f(0.1)=0.1\;\;\;\;$ You can develop the Schroeder function at either fixed point. 0.1 is a repelling fixed point, -0.1 is an attracting fixed point.
Now consider the first half of the solution to the Op's function as $g(z)=f^{o z}(0)\;\;\;\;g(z+1)=f(g(z))$, where $f^{o z}(0)$ is generated from the +0.1 fixed point using the inverse Abel function $\alpha^{-1}(z)$, and the Abel function is generated from standard Schroeder function solution around the +0.1 fixed point. $g(z)$ is a complex periodic function with a period of $\frac{2\pi i}{\ln(\lambda)}$ where $\lambda$ is the multiplier at the fixed point. For the fixed point of 0.1, $\lambda=\frac{6}{5}$ since $f(0.1+x)=0.1+\frac{6x}{5}+x^2$, and the period is $\approx 34.462i$, so the $g(z)$ function is imaginary periodic, and is generated from the Schroeder equation for $y \mapsto \frac{6}{5}y+y^2;\;\; y=z-0.1\;\;\;\;$ Also, $g(z)$ turns out to be entire. Here is the periodic Fourier series form for $g(z)$.
$$g(z) = 0.1 + \sum_{n=1}^{\infty} a_n \left( \frac{6}{5} \right) ^ {nz} $$
Now, consider another function $h(z)$, which is generated from the other -0.1 fixed point in much the same way. It turns out that $h(z)=g(z+\theta(z))$, where $\theta(z)$ is a 1-cyclic periodic function, $\theta(z+1)=\theta(z)$. But first, back to $h(z)$.
$h(z)=f^{o z}(0)\;\;\;\;h(z+1)=f(h(z))$, but this time we are using the other fixed point of -0.1, which is an attracting fixed point. $h(z)$ is generated from the -0.1 fixed point using the inverse Abel function $\alpha^{-1}(z)$, and the Abel function is generated from standard Schroeder function solution around the -0.1 fixed point. $h(z)$ is a complex periodic function with a period of $\frac{2\pi i}{\ln(\lambda)}$ where $\lambda$ is the multiplier at the fixed point. For the fixed point of -0.1, $\lambda=\frac{4}{5}$ since $f(-0.1+x)=-0.1+\frac{4x}{5}+x^2$, and the period is $\approx 28.157i$ The $h(z)$ function is imaginary periodic, and is generated from the Schroeder equation for $y\mapsto \frac{4}{5}y+y^2;\;\; y=z+0.1\;\;\;\;$ $h(z)$ has square root branches where $h(z)=-0.5$. These singularities occur at half the imaginary period of $h(z)$, plus any multiple of the period; one of them is near $z\approx 4.0831 \pm 14.0788i$. $h(z)$ is analytic to the right of this singularity, and where also where $|\Im(z)|<\approx 14.0788i$
$$h(z) = -0.1 + \sum_{n=1}^{\infty} b_n \left( \frac{4}{5} \right) ^ {nz} $$
Both $h(z)$ and $g(z)$ are real valued at the real axis; both have the same defining iteration equations, and for integer values of z, $h(z)=g(z)$. They disagree very slightly on fractional iterations, which is what leads to $\theta(z)$. Both h(z) and g(z) have a limiting value of -0.1 as real z gets arbitrarily large positive, and a limiting value of 0.1 as z real z gets arbitrarily large negative. g(z) is well defined for complex values of z if $\Re(z)$ is a large negative number, where $g(z)\approx 0.1+1.2^{z+k}$, and is imaginary periodic, with values cycling around 0.1 with an imaginary period of $\approx 34.462i$. At half the imaginary period, $g(z)$ grows from the 0.1 repelling fixed point and gets arbitrarily large. $h(z)$ is well defined in the complex plane if $\Re(z)$ is a large positive number, where $h(z)\approx -0.1$, and $h(z)$ is also imaginary periodic, and $h(z) \approx -0.1 + 0.8^{z+k}$ is cycling around -0.1 with a different imaginary period of $\approx 28.157i$
So, it turns out that there is a small one cyclic periodic function, $f(z)=\theta(z)$. $\theta(z)$ is analytic at the real axis, and has a 1-cyclic singularity at $z \approx n + 0.0831 \pm 14.0788i$, where the singularities of $h(z)$ are. And, an example solution to the Op's question becomes:
$$h(z)= g(z+\theta(z))$$
Here is a picture of the iterated function $f^{o z}(0)$ at the real axis, from -25 to +25, where it gradually goes from +0.1 to -0.1, where $f(z)=z^2+z-0.01$ Integer values of $f^{o z}$ can be trivially generated by iterating $z \mapsto z^2+z-0.01$ or iterating the inverse function, $f^{-1}(z)$ which is $z \mapsto \sqrt{z+0.26}-0.5$. For fractional iterations of $f^{o z}$, there are two very slightly different solutions, depending on whether we develop the iterated function from the attracting fixed point of -0.1 at $+\infty$ or the repelling fixed point of +0.1 at $-\infty$. These two solutions for $f^{o z}(0)$ can be generated from the two formal Schroeder equations for the two fixed points, although the formal solution needs to be scaled so that g(0)=0 and h(0)=0. I can add later.
Here is a graph of the tiny real valued 1-cyclic $\theta(z)$ function at the real axis, from -1 to +1. As you can see, its magnitude is $|\theta(z)| \lt 8 \cdot 10^{-40}$
Here is a graph of $\left(g(z)-0.1\right)$ at $\;\;\Re(z)=-25$. I used the iteration function $y \mapsto \frac{6y}{5}+y^2$ where $z=y+0.1$ and $z \mapsto z+z^2-0.01$. Here, $g(z)$ has an imaginary period of $\frac{2\pi i}{\ln(1.2)}\approx 34.462i$. The graph goes from $z=-25-18i$ to $z=-25+18i\;\;$ Red is imaginary and magenta is real. A similar graph could be generated for $h(z)+0.1$ at $\Re(z)=+25$, but the imaginary period would be $\approx 28.158i$.