why only impulse signal for convolution?
Because the impulse response completely characterizes an LTI system. The reason is an arbitrary input $x(t)$ can be written as an infinite sum of time shifted and scaled Delta functions:

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

or alternatively, using the convolution operator, $$x(t) = x(t)*\delta(t)$$

Properties of linearity and time invariance lead to the following sequence of input-output pairs:


$\delta(t) \longrightarrow h(t)$
Time invariance: $\delta(t-\tau) \longrightarrow h(t-\tau)$ 
Linearity: $x(\tau)\delta(t-\tau) \longrightarrow x(\tau)h(t-\tau)$
Linearity: $\int_{-\infty}^{+\infty}x(\tau)\delta(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau$


The last input-output pair can be written as $$x(t) \longrightarrow y(t) = x(t)*h(t)$$

This way, any output can be computed using the input $x(t)$, the impulse response $h(t)$, and the convolution integral (or sum, in discrete time).

@a concerned citizen gave you some other hints regarding your question.

On the contrary, if you use the unit step function, you get

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

or alternatively, using the convolution operator, $$\int_{-\infty}^t x(t) = x(t)*u(t)$$

Once again, just for comparison, you get the following sequence of input-output pairs:


$u(t) = \int_{-\infty}^t \delta(\tau)d\tau \longrightarrow s(t) = \int_{-\infty}^t h(\tau)d\tau$
Time invariance: $u(t-\tau) \longrightarrow s(t-\tau)$ 
Linearity: $x(\tau)u(t-\tau) \longrightarrow x(\tau)s(t-\tau)$
Linearity: $\int_{-\infty}^{+\infty}x(\tau)u(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)s(t-\tau)d\tau$


where $s(t)$ is the step response, the system output when the unit step is presented at the input. The last input-output pair can be written as $$\int_{-\infty}^t x(\tau)d\tau \longrightarrow x(t)*s(t) = x(t)*u(t)*h(t)$$

The latter can be written as $$(x(t)*h(t))*u(t) = y(t)*u(t) = \int_{-\infty}^t y(\tau)d\tau$$

So after all, what you get at the output is the integral of the output you would get using the delta decomposition of an input signal. That is, your output must be differentiated to obtain the actual system output for a given $x(t)$.