You don't say what the morphisms in $C$ are supposed to be. I'll take them to be isotopy classes of orientation-preserving diffeomorphisms. Let $\mathbb{N}$ be the category with object set $\mathbb{N}$ and only identity morphisms. Then the genus gives a functor $\pi\colon C\to\mathbb{N}$, and by choosing a surface $X_g$ of genus $g$ for all $g\geq 0$ we get a functor $\sigma\colon\mathbb{N}\to C$. We also have $F_0\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ given by $F_0(n,m)=2nm$, and then $F=\sigma\circ F_0\circ \pi^2$ is a functor of the type that you asked for.

Although this seems very degenerate, it would not surprise me if there were few other possibilities. Indeed, if we put $G_n=C(X_n,X_n)$, then $F$ will give group homomorphisms $F_{n,m}\colon G_n\times G_m\to G_{2nm}$. This means that $F_{n,m}(g,1)$ must commute with $F_{n,m}(1,h)$ for all $h$, which is already quite restrictive. There are known presentations of the groups $G_n$ in terms of Dehn twists, and one could try to use those to analyse the possibilities, perhaps starting with $n=m=1$, where $G_1=SL_2(\mathbb{Z})$.

(I am assuming here that you deliberately wrote $b_1(F(S,S'))=b_1(S)b_1(S')$ and so $\text{genus}(F(S,S'))=2\text{genus}(S)\text{genus}(S')$. If in fact you meant $\text{genus}(F(S,S'))=\text{genus}(S)\text{genus}(S')$ then the answer would be similar but the case $n=m=1$ would be easier to analyse as you would then be looking for homomorphisms $SL_2(\mathbb{Z})\times SL_2(\mathbb{Z})\to SL_2(\mathbb{Z})$. Here of course you have the two projections and there are also various things you can do with the abelianization map $SL_2(\mathbb{Z})\to\mathbb{Z}/12$.)

You don't say what the morphisms in $C$ are supposed to be. I'll take them to be isotopy classes of orientation-preserving diffeomorphisms. Let $\mathbb{N}$ be the category with object set $\mathbb{N}$ and only identity morphisms. Then the genus gives a functor $\pi\colon C\to\mathbb{N}$, and by choosing a surface $X_g$ of genus $g$ for all $g\geq 0$ we get a functor $\sigma\colon\mathbb{N}\to C$. We also have $F_0\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ given by $F_0(n,m)=2nm$, and then $F=\sigma\circ F_0\circ \pi^2$ is a functor of the type that you asked for.

Although this seems very degenerate, it would not surprise me if there were few other possibilities. Indeed, if we put $G_n=C(X_n,X_n)$, then $F$ will give group homomorphisms $F_{n,m}\colon G_n\times G_m\to G_{2nm}$. This means that $F_{n,m}(g,1)$ must commute with $F_{n,m}(1,h)$ for all $h$, which is already quite restrictive. There are known presentations of the groups $G_n$ in terms of Dehn twists, and one could try to use those to analyse the possibilities, perhaps starting with $n=m=1$, where $G_1=SL_2(\mathbb{Z})$.

(I am assuming here that you deliberately wrote $b_1(F(S,S'))=b_1(S)b_1(S')$ and so $\text{genus}(F(S,S'))=2\text{genus}(S)\text{genus}(S')$. If in fact you meant $\text{genus}(F(S,S'))=\text{genus}(S)\text{genus}(S')$ then the answer would be similar but the case $n=m=1$ would be easier to analyse as you would then be looking for homomorphisms $SL_2(\mathbb{Z})\times SL_2(\mathbb{Z})\to SL_2(\mathbb{Z})$. Here of course you have the two projections and there are also various things you can do with the abelianization map $SL_2(\mathbb{Z})\to\mathbb{Z}/12$.)

[UPDATE] If $F$ is supposed to be functorial for all continuous maps, then in particular it is functorial for homeomorphisms. If $F$ is also a continuous functor then it will send isotopic homeomorphisms to isotopic homeomorphisms. The group of isotopy classes of homeomorphisms is isomorphic to the group of smooth isotopy classes of diffeomorphisms, so we we get back to the framework described above, apart from the orientation issue, which does not make much difference. However, if we want functoriality for arbitrary continuous maps then we no longer have a functor $\pi\colon C\to\mathbb{N}$. In this context I think it is very unlikely that there are any functors $F$ at all.