Electronic – Convolution with two unbounded signals

So I have a convolution problem. I have two signals:

I need to find the output, y(t), through graphical convolution.

So I set the convolution integral up like this:
$$
\int h(\tau)*x(t-\tau)d\tau
$$

I set it up this way because I believed it easier to 'flip and slide' the exponential function than the unit step function. So, then I substituted in \$ h(\tau)\$ and \$x(t-\tau)\$ like so:
$$
\int_{t-\inf}^{0} 1*e^{t-\tau}d\tau
$$
This then gave me:
$$
-e^{t-\tau}|_{t-\inf}^0
$$
Which equals \$-e^t\$. This is fine, and acceptable in my opinion. However, when moving to the next section, where the exponential is partially outside of the unit step function, I get issues. The equation I used here was this:
$$
\int_{-\inf}^0e^{t-\tau}d\tau + \int_{t-\inf}^te^{t-\tau}d\tau
$$
This includes the previous section from infinity to 0, and adds the new section, where the exponential has now moved partially outside of the step function, being bound by \$t-\inf\$, as before, and \$t\$. This then gives me this:
$$
-e^{t-\tau}|_{-\inf}^0 + -e^{-t-\tau}|_{t-\inf}^t
$$
This is where I start to have problems. \$-e^{t-\tau}|_{-\inf}^0\$ evaluates to:
$$
-e^t + e^{t+\inf}
$$
This is, of course, an issue, as \$e^{t+\inf}\$ is infinite, and cannot be used as it doesn't converge.

Is there a different way I can graphically convolve these two signals without running into issues with infinity?

Now, I have done this problem using \$h(t-\tau)\$ and \$x(\tau)\$. This runs into a similar issue. For the first section, where the step function is partially outside of the exponential, I get this:
$$
\int_{-\inf}^0e^-td\tau
$$
That evaluates to this:
$$
-te^{-t}
$$
Fine. Good. Now, when I do the next section, where the step function has passed \$\tau = 0\$, and now the limit of the step function is going along the curve of the exponential, I have this:
$$
\int_{-\inf}^0e^{-t}d\tau + \int_0^te^{-t}d\tau
$$
This evaluates to \$\tau e^{-t}|_{-\inf}^0+\tau e^{-t}|_0^t\$, or \$\inf +te^{-t}\$.

So I run into a similar issue to what I had had before.

So I decided to do this convolution using Laplace transforms.
Since:
$$
x(t) = e^{-t}
$$
and
$$
h(t) = u(-t)
$$
I found these transformations:
$$
H(s) = \frac1s; X(x) = \frac{1}{s+1}
$$
Multiplying these, I find:
$$
Y(s) = \frac{1}{s^2+1}
$$

Applying an inverse Laplace Transform, I get:
$$
y(t) = 1 – e^{-t}
$$