Setting
Let us regard the Hilbert space $L^2(0,1)$ and the $C_0$-semigroup $(T(t))_{t\geq 0}$ defined by $$ T(t):\left\{ \begin{array}{rml} L^2(0,1) & \to & L^2(0,1), \\ f &\mapsto &\left(x \mapsto \begin{cases} f(x+t), & \text{if}\; x+t<1\\ 0, & \text{else} \end{cases} \right). \end{array} \right. $$ It is easy to verify that this is indeed a $C_0$-semigroup. Therefore, the mapping $t \mapsto T(t)f$ is a continuous mapping from $L^2(0,1)$ to $L^2(0,1)$. Consequently the $L^2(0,1)$-valued integral $$ g := \int_0^1 T(t)f \,\mathrm{d}t $$ exists.
Question
In order to get some information about the behavior of $g$ it would be nice to regard $g$ as a parameter integral. Hence I am interested in the following equality $$ g(x) = \Big(\int_0^1 T(t)f \,\mathrm{d}t\Big) (x)\stackrel{?}{=} \int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t . $$ Or with a different notation $$ g = \int_0^1 \big(x \mapsto \big(T(t)f\big)(x) \big)\,\mathrm{d}t \stackrel{?}{=} \Big(x\mapsto\int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t\Big) . $$ The evaluation mapping is neither continuous nor well-defined on $L^2$. So I think it is not trivial to justify this step.
It seems quite common to evaluate such $L^2(0,1)$-valued integrals by interpreting it as a parameter integral, so I guess that there is a theorem which justifies that. It would be really great if someone had a reference.
Solution for this special case
In this particular case I think I have a solution. I know that every convergent sequence in $L^2$ has a subsequence which converges even point-wise a.e.. Since $$ g_n := x\mapsto \sum_{i=1}^{n} \frac{1}{n} \Big(T\Big(\frac{i}{n}\Big)f\Big)(x) $$ converges to $g$ and every subsequence of $g_n(x)$ converges in $\mathbb{R}$ to the same limit for a.e. $x\in (0,1)$, the point-wise limit of $g_n$ has to coincide with $g$ a.e..
$\newcommand{\d}{\,\mathrm{d}}$ $$ g(x) = \int_{0}^{1}(T(t)f)(x)\d t=\int_{0}^{1-x}f(x+t)\d t=\int_{x}^{1}f(y)\d y. $$ You may use a Riemann integral for the function $t\mapsto T(t)f$ because it is a continuous function with respect to the $L^2$ norm. The resulting integral is in $L^2$. So, the following holds for al $h\in L^2$ because it holds for the Riemann sums in $t$: $$ \langle g,h\rangle = \Big\langle \int_{0}^{1}T(t)f \d t,h\Big\rangle \\ = \int_{0}^{1}\langle T(t)f,h\rangle \d t $$ Then the above may be written as $$ \langle g,h\rangle =\int_{0}^{1}\int_{0}^{1}\chi_{[0,1]}(x+t)f(x+t)h(x)\d x \d t \\ =\int_{0}^{1}\int_{0}^{1}\chi_{[0,1]}(x+t)f(x+t)\d t\, h(x)\d x \\ =\int_{0}^{1}\int_{0}^{1-t}f(x+t)\d t\, h(x)\d x \\ =\Big\langle \int_{0}^{1-t}f(x+t)\d t,h\Big\rangle. $$ This holds for all $h\in L^2$, which gives $$ g(x) = \int_{0}^{1-x}f(x+t)\d t,\;\; a.e. x. $$