Suppose that we have two Binomial Likelihood functions, i.e.
$$L_{1} = \binom{N}{x}p^{x}(1-p)^{N-x}$$
$$L_{2} = \binom{M}{y}p^{y}(1-p)^{N-y}$$
The total likelihood is the product of those two likelihood functions
$$L_{total}= L_{1}\times L_{2} =\binom{N}{x}p^{x}(1-p)^{N-x} \binom{M}{y}p^{y}(1-p)^{N-y}$$ $$=\binom{N}{x}\binom{M}{y}p^{x+y}(1-p)^{(N+M)-(x+y)}$$
We also have knowledge of the number of counts $x$ and $y$ and the sum $N+M$. However, we do not know the values of $N$ and $M$ individually in order to be able to calculate the binomial coefficients $\binom{N}{x}$ and $\binom{M}{y}$.
Is there a way to tackle this problem, of the absence of knowledge of $N$ and $M$, in order to be able to use the likelihood function $L_{total}$?
Unfortunately $L_{total}$ you seek is not independent of exact values of $M$ and $N$. You can check this easily by plugging in some values.
Case 1: $M=5$, $N=5$, $p=0.5$, $x=2$, $y=3$
Result: $L_1 = 0.3125, L_2= 0.3125, L_{total} = 0.09765625$
Case 2: $M=2$, $N=8$, $p=0.5$, $x=2$, $y=3$
Result: $L_1 = 0.25, L_2= 0.21875, L_{total} = 0.0546875$