So all this is about the theory of distributions. In this theory, indeed, a generalized function $f$ is a linear form on a space of test functions. It is thus defined by its action on tests functions $\varphi\in C^\infty_c$, that one can indeed write $\langle f,\varphi\rangle$. The fact that its generalizes functions comes from the fact that for each locally integrable function, one associate the distribution defined by
$$
\langle f,\varphi\rangle = \int_{\Bbb R} f(x)\,\varphi(x)\,\mathrm d x.
$$
But in general, $\langle f,\varphi\rangle$ might not be possible to write as the integral of a product of functions. For example, the Dirac delta $\delta_0$ is defined by $\langle \delta_0,\varphi\rangle = \varphi(0)$.
Now, one can extend the usual operations by getting looking at what happens when $f$ is a nice function. For example the derivative of a distribution is defined by $\langle f',\varphi\rangle = -\langle f,\varphi'\rangle$, because for nice functions $f$, this relation is true. Similarly, to define the convolution of a distribution with a test function, one can look at the case of nice functions $f$ for which
$$
f*\varphi(y) = \int_{\Bbb R} f(x)\,\varphi(y-x)\,\mathrm d x = \langle f,\varphi(y-\cdot)\rangle
$$
and so one can define more generally $f*\varphi(y) := \langle f,\varphi(y-\cdot)\rangle$ even if $f$ is not a locally integrable function. In your case, this gives indeed
$$
f*\varphi(y) = \int_{\Bbb R} f(x)\left(\varphi(y-x)-\varphi(y)\mathbf{1}_{|x|<1}\right)\mathrm d x
$$
Now, this is a classical integral. You can do the change of variable $x\mapsto y-x$, but it will give you
$$
f*\varphi(y) = \int_{\Bbb R} f(y-x)\left(\varphi(x)-\varphi(y)\mathbf{1}_{|x-y|<1}\right)\mathrm d x,
$$
which is a bit different from your last expression. And indeed, your last expression cannot be right, since the difference $\varphi(x)-\varphi(0)$ is initially here to compensate the singularity of $f$ at $0$. But in your last expression, the singularity of $f$ occurs at $x=y$, but $\varphi(y)-\varphi(0)$ is in general not small. At the contrary, in my last expression $\varphi(x)-\varphi(y)$ is indeed small when $x\to y$.
Remarks:
If the test functions are in $C^\infty_c$ then you are dealing with distributions. Tempered distributions are the particular case of distributions where the test functions can be chosen in $C^\infty$ with fast decay at infinity. It is indeed the case for your particular $f$.
This regularization does not come from nowhere. Indeed, you get these kind of distributions by taking derivatives of functions in the sense of distributions (that is using the rule $\langle f',\varphi\rangle = -\langle f,\varphi'\rangle$). You can look for example at this post where I proved (take $d=1$) that the derivative of $u(x) = \mathrm{sgn}(x) \ln(|x|)$ is the distribution defined by
$$
\langle u',\varphi\rangle = \int_{\Bbb R} \frac{\varphi(x)-\varphi(0)\mathbf{1}_{|x|<1}}{|x|}\,\mathrm{d}x.
$$
I suppose you get something close to your distribution by the same proof and looking at the derivative of the function $g(x) = \mathrm{sgn}(x) \ln(1-e^{-|x|})$.
