You are asking for the distribution of the sum of independent random variables, one uniformly distributed, the other gamma. (the Q do not specify independence, but it seems to be assumed).
So let $Y=U+X$ where $U \sim \mathcal{U}(0, b), b>0$ and $X\sim \mathcal{Gamma}(\alpha, \beta), \alpha>0, \beta>0$ where $\beta$ is the rate. Write the respective densities as
$$ g(u)=\frac1b \mathcal{I}(0\le u \le b) $$ and
$$ h(x)=\frac{\beta^\alpha}{\Gamma(\alpha) } x^{\alpha-1} e^{-\beta x} \cdot
\mathcal{I}(x>0) $$
Then we find the density of $Y$ with the convolution integral
$$ f(y) = \int g(u) h(y-u) \; du = \\
\int_0^b \frac1b \frac{\beta^\alpha}{\Gamma(\alpha) }(y-u)^{\alpha-1} e^{-\beta (y-u)} \cdot \mathcal{I}(x>0) \; du
$$ These can be solved separately for the two cases $y \gtrless b$, by a change of variable and recognizing in the integrand the gamma density, resulting in
$$ f(y) = \begin{cases} \frac1b F_{\alpha, \beta}(y) & y \le b \\
\frac1b \left\{ F_{\alpha, \beta}(y) - F_{\alpha, \beta}(y-b) \right\} & y > b
\end{cases} $$
where $F_{\alpha, \beta}$ is the cdf (cumulative distribution function) of the Gamma distribution as above.
Edit
Looking through the answer, it is clear that the gamma distributions special form do not play a role at all. So, whatever the distribution of $X$, the answer has the same form. For the sake of generality, let us show this. So, notation as above, but with $X \sim F$, whatever cdf, without restrictions.
$$ \DeclareMathOperator{\P}{\mathbb{P}}
\P(Y \le y) = \P(U+X\le y) =\int \P(U\le y-X \mid X=x)\; dF(x) =\\
\int F_U(y-x)\; dF(x) $$ differentiating under the integral gives
$$ f(y)= \int g_U(y-x)\; dF(x) $$ and using the uniform distribution of $U$
$$ f(y)= \frac1b \int_{y-b}^y dF(x) = \frac1b \left\{ F_X(y)-F_X(y-b) \right\} $$
