Lets consider a continuous-time LTI system where $x(t) =\delta(t)$, then we have 
\begin{align*}
x \to &\fbox{H} \to y \\
\delta(t) \to &\fbox{H} \to h(t) \\
\end{align*}

where $h(t)$ represents our impulse response. Note that if we mention ``impulse response'', then it is implied that a given system is LTI. We can demonstrate this property here: 

$$ \alpha\delta(t-T) \to \fbox{H} \to \alpha h(t-T) $$

We know from the sifting property, we can write any signal as the integral of the product of itself and the Dirac Delta:

$$ x(t) = \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau$$

This looks very similar to the DT counterpart, where we write a signal as a linear combination of impulses:

\begin{align*}
x(n) &= \sum \limits_{m=-\infty}^{\infty}x(m)\delta(n-m) \\
y(n) &= \sum \limits_{m=-\infty}^{\infty}x(m)h(n-m) = (x*h)(n)\\
\end{align*}

Just as we wrote signals in DT as sums of weighted delta functions, in CT we can write the convolution integral:

\begin{nicebox}
$$y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau = (x*h)(t) $$
\end{nicebox}

The convolution operator still has its commutativity property, which we can demonstrate by a substitution:

\begin{align*}
y(t) &= \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau = (x*h)(t)\\
y(t) &= \int_{\infty}^{-\infty}x(t-\mu)h(\mu)\left(-d\mu\right) \quad \mbox{let } \mu=t-\tau, \mbox{limits have switched}\\
y(t) &= \int_{-\infty}^{\infty}h(\mu)x(t-\mu)d\mu \quad \mbox{absorb negative by switching limits again }\\
y(t) &= \int_{-\infty}^{\infty}h(\mu)x(t-\mu)d\mu = (h*x)(t)\\
\end{align*}

LTI Systems and their impulse responses