The standard definition of expected value $E(x)$ of a continuous distribution $\rho(x)$ is given by $$ E(x) = \int_\mathbb{R} x \rho(x) dx $$
Expected value is exactly what you might think it means intuitively: the return you can expect for some kind of action, like how many questions you might get right if you guess on a multiple choice test.
This is fine whenever $x$ has no units, such as tossing an infinite dimensional coin. Otherwise, however, this definition has incorrect units. For instance,
Suppose $\rho(x)$ is a distribution describing the probability of a ball landing at point $x$, so that $E(x)$ is the mean value of position landing. We expec it to have units of position, obviously. However, $\int x \rho(x) dx$ has units of position^2.
Where is the mistake? It could be in the definition of $E(x)$. Replacing it with $$ E(x) = \dfrac{\int x \rho(x) dx}{\int \rho(x) dx} $$ Now, dimensional analysis matches our expectations, and since $\int \rho(x) dx = 1$ (with units!) anyway, the original expression is not lost. Two piece of evidence in favor of this definition are presented below.
Quantum mechanics:
In QM, $\psi^2(x)$ is a function describing the probability of finding a particle at point $x$, and is expected value is defined as $$ E(x) = \dfrac{\int x \psi^2(x) dx}{\int \psi^2(x) dx} $$ Even though $\psi$ is in $L^2$, $\int \psi^2(x) dx$ need not be $1$. If it is, though, the original expression for $E(x)$ is recovered (with correct units).
Weighted arithmetic mean:
The definition of $E(x)$ many times comes from a generalization weigthed arithmetic mean to the continuous case. $$ \bar{x} = \frac{\sum w_i x_i}{\sum w_i} $$ From this point of view, the proposed expression is more similar to the above than the standard. It just happens that if the weigths describe probabilities, then $\sum w_i = 1$, again.
I ask is the proposed definition of $E(x)$ correct? Otherwise, how does the original definition of $E(x)$ yields correct units?
The mistake lies in the fact that you think $\rho(x)$ is unit-less. It is in fact a probability density, and a careful inspection of its definition tells you that it has units of $[\mbox{Probability}] / [x] = [x]^{-1}$ This is because, by definition, $\rho(x)\Delta x$, for small $\Delta x$ gives you the approximate probability of falling into a bin $[x,x+\Delta x]$. Probability on its own is unit less (think of it as a weight), but I've included to emphasize density.