I have a feeling that uncountable products in $MBool$ do not exist in general. One thing is for sure: if $MBool$ has uncountable products, then the forgetful functor $MBool \to Bool$ cannot preserve them (whence the forgetful functor couldn't have a left adjoint.)

Indeed, if $2$ is the 2-element Boolean algebra with its unique measure $\mu$. If the uncountable product $(2, \mu)^{\omega_1}$ existed in $MBool$, where $\omega_1$ is the first uncountable ordinal, and if this product is calculated as it would be in $Bool$ (which is the set-theoretic infinite product), then we could construct an uncountable strictly decreasing chain of elements (uncountable tuples)

$$(1, 1, 1, \ldots) > (0, 1, 1, \ldots) > (0, 0, 1, \ldots) > \ldots$$

and there would have to be a corresponding uncountable strictly decreasing chain of positive real numbers

$$\mu(1, 1, 1, \ldots) > \mu(0, 1, 1, \ldots) > \mu(0, 0, 1, \ldots) > \ldots$$

by the axioms on metric Boolean algebras. This is impossible by cofinality considerations.

Regardless of whether there are infinite products in $MBool$, there is a trivial reason why the forgetful functor $U: MBool \to Bool$ cannot have a left adjoint $F$ under the axioms given for metric Boolean algebras. Suppose WLOG that $\mu(1) = 1$ where the $1$ on the left is the top element of the putative free metric Boolean algebra $FB$. Then define a different metric $\mu'$ on $UFB$ by $\mu'(b) = r\mu(b)$ where $r > 1$, giving a different metric Boolean algebra $B'$. Then there is no *nonexpansive* metric Boolean algebra map $FB \to B'$ which extends the Boolean algebra embedding $i: B \to UB'$ along $i: B \to UFB$.

One might think this trivial objection could be remedied by adding an extra axiom like $\mu(1) = 1$ (so we are working with probability measures in a sense), but it seems highly doubtful even in that case that a left adjoint out of $Bool$ would exist. As noted above, it could only exist if $MBool$ lacked many limits like uncountable products, and this would rule out a straightforward application of an adjoint functor theorem.

I am not sure about a left adjoint out of $Met$.