The convolution of $f$ and $g$ is defined as $ (f * g )(t) \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau) \, d\tau $.
Let's say that $f(t)$ and $g(t)$ have units of, say, meters and Hertz, respectively. What units does the convolution $(f * g )(t)$ have?
Let $X$ and $Y$ be independent random variables with densities $f, g$ respectively. Then the convolution of $f$ and $g$ $$ (f * g )(t) \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau) \, d\tau $$ is the density function of $X+Y$. For this to make dimensional meaning, we must assume that $X$ and $Y$ are measured with the same units of measurement, it could for instance be m/s (meters per second). Then the unit of measurement of the probability density function is probability per (m/s). Since probability is a pure number we can write this as 1/(m/s). Let us make this more general by writing u for whatever common unit of measurement of $X$ and $Y$, then the densities $f,g$ has units 1/u.
We can represent the convolution integral above with a Riemann sum: $$ = \sum_i f(\tau_i) g(t-\tau_i) (\tau_i -\tau_{i-1}) $$ The variable $\tau$ above clearly has unit u. So the unit of measurement of each term in the Riemann sum has unit $$ \frac{1}{\text{u}}\cdot \frac{1}{\text{u}}\cdot\text{u} = \frac{1}{\text{u}} $$ showing that the density function obtained by convolving $f$ and $g$ has the same unit as $f$ and $g$.
