Impulse response (general form for linear systems as two-variables function $h(t,\tau)$) applied to Time-invariant systems If a system is linear, then the operator $S$ mapping input signals into output signals - i.e. $y(t)=S\{u(t)\}(t)$ - is the integral of the input weighted by the impulse response:
$$ y(t) = \int_{-\infty}^{+\infty} h(t,\tau) u(\tau) \mathrm{d}\tau $$
where $h(t,\tau):=S\{\delta(t-\tau)\}$, response of the system to an impulse in $\tau$, is a function of two variables: the second one $\tau$ refers to the time at which the input is applied while the first one $t$ refers to the time at which the output is observed.

If this system is also time-invariant, the system operator commutes with the time shift operator:
$$S\{u(t-T)\}(t) = S\{u(t)\}(t-T) $$
that is, time shifting the input, a time shifted version of the output is produced.
Linear time invariant systems have an impulse response function which depends upon a single variable, the difference between $t$ and $\tau$:
$$h(t,\tau)=\tilde h(t-\tau)$$ where the $\sim$ symbol is used just to underline that $h$ and $\tilde h$ cannot be the same, because they have a different number of arguments.
How can I show that the impulse response can be expressed as a function depending only on $t-\tau$ starting from the definition of time invariance?
If the response to $x(t)$ is given by

$$y(t)=\int_{-\infty}^{\infty}h(t,\tau)x(\tau)d\tau\tag{1}$$

then the response to $x(t-T)$ is

$$\tilde{y}_T(t)=\int_{-\infty}^{\infty}h(t,\tau)x(\tau-T)d\tau=\int_{-\infty}^{\infty}h(t,\tau+T)x(\tau)d\tau\tag{2}$$

If the system is time-invariant we require

$$\tilde{y}_T(t)\stackrel{!}{=}y(t-T)=\int_{-\infty}^{\infty}h(t-T,\tau)x(\tau)d\tau\tag{3}$$

For $(2)$ and $(3)$ to be equal for any $x(t)$ we must have

$$h(t,\tau+T)\stackrel{!}{=}h(t-T,\tau)\tag{4}$$

which is equivalent to

$$h(t+T,\tau+T)\stackrel{!}{=}h(t,\tau)\tag{5}$$

From $(5)$ it is clear that for a linear and time-invariant system the value of $h(t,\tau)$ only depends on the difference of its arguments, and not on their individual values. Consequently, the impulse response can be rewritten as a function of $t-\tau$:

$$h(t,\tau)=\tilde{h}(t-\tau)\tag{5}$$